Partial Differential Equation
P P Savani University
March 16, 2022
(P P Savani University)
Partial Differential Equation
March 16, 2022
1 / 55
Introduction
Introduction
Let z = f (x, y) be the function where x and y are the independent variables and z,
the dependent variable. Consider following notations:
∂z
= p,
∂x
(P P Savani University)
∂z
= q,
∂y
∂ 2z
= r,
∂x2
∂ 2z
= s,
∂x∂y
Partial Differential Equation
∂ 2z
=t
∂y 2
March 16, 2022
2 / 55
Formation of First and Second order equations
Formation of PDE
Unlike the case of ODE which arise from the elimination of arbitrary constants;
the PDE can be formed either by the elimination of arbitrary constants or by the
elimination of arbitrary functions from a relation involving three or more variables.
(P P Savani University)
Partial Differential Equation
March 16, 2022
3 / 55
Formation of First and Second order equations
Formation of PDE
Unlike the case of ODE which arise from the elimination of arbitrary constants;
the PDE can be formed either by the elimination of arbitrary constants or by the
elimination of arbitrary functions from a relation involving three or more variables.
Example: Form PDE for z = ax + by + ab.
(P P Savani University)
Partial Differential Equation
March 16, 2022
3 / 55
Formation of First and Second order equations
Formation of PDE
Unlike the case of ODE which arise from the elimination of arbitrary constants;
the PDE can be formed either by the elimination of arbitrary constants or by the
elimination of arbitrary functions from a relation involving three or more variables.
Example: Form PDE for z = ax + by + ab.
Solution: Differentiating partially w.r.t x and y,
(P P Savani University)
Partial Differential Equation
March 16, 2022
3 / 55
Formation of First and Second order equations
Formation of PDE
Unlike the case of ODE which arise from the elimination of arbitrary constants;
the PDE can be formed either by the elimination of arbitrary constants or by the
elimination of arbitrary functions from a relation involving three or more variables.
Example: Form PDE for z = ax + by + ab.
Solution: Differentiating partially w.r.t x and y,
∂z
= a i.e., p = a
∂x
(P P Savani University)
Partial Differential Equation
March 16, 2022
3 / 55
Formation of First and Second order equations
Formation of PDE
Unlike the case of ODE which arise from the elimination of arbitrary constants;
the PDE can be formed either by the elimination of arbitrary constants or by the
elimination of arbitrary functions from a relation involving three or more variables.
Example: Form PDE for z = ax + by + ab.
Solution: Differentiating partially w.r.t x and y,
∂z
= a i.e., p = a
∂x
∂z
=b
∂y
(P P Savani University)
i.e.,
q=b
Partial Differential Equation
March 16, 2022
3 / 55
Formation of First and Second order equations
Formation of PDE
Unlike the case of ODE which arise from the elimination of arbitrary constants;
the PDE can be formed either by the elimination of arbitrary constants or by the
elimination of arbitrary functions from a relation involving three or more variables.
Example: Form PDE for z = ax + by + ab.
Solution: Differentiating partially w.r.t x and y,
∂z
= a i.e., p = a
∂x
∂z
= b i.e., q = b
∂y
Substituting the values in the given equation, z = px + qy + pq is the required partial
differential equation of first order.
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
3 / 55
Formation of First and Second order equations
Example: Eliminate the arbitrary constants a & b from (x − a)2 + (y − b)2 + z 2 = 1
and form the PDE.
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
4 / 55
Formation of First and Second order equations
Example: Eliminate the arbitrary constants a & b from (x − a)2 + (y − b)2 + z 2 = 1
and form the PDE.
Solution: Differentiating partially w.r.t. x and y,
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
4 / 55
Formation of First and Second order equations
Example: Eliminate the arbitrary constants a & b from (x − a)2 + (y − b)2 + z 2 = 1
and form the PDE.
Solution: Differentiating partially w.r.t. x and y,
2(x − a) + 2zp = 0
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
4 / 55
Formation of First and Second order equations
Example: Eliminate the arbitrary constants a & b from (x − a)2 + (y − b)2 + z 2 = 1
and form the PDE.
Solution: Differentiating partially w.r.t. x and y,
2(x − a) + 2zp = 0
2(y − b) + 2zq = 0
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
4 / 55
Formation of First and Second order equations
Example: Eliminate the arbitrary constants a & b from (x − a)2 + (y − b)2 + z 2 = 1
and form the PDE.
Solution: Differentiating partially w.r.t. x and y,
2(x − a) + 2zp = 0
2(y − b) + 2zq = 0
Substituting the values in the given equation,
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
4 / 55
Formation of First and Second order equations
Example: Eliminate the arbitrary constants a & b from (x − a)2 + (y − b)2 + z 2 = 1
and form the PDE.
Solution: Differentiating partially w.r.t. x and y,
2(x − a) + 2zp = 0
2(y − b) + 2zq = 0
Substituting the values in the given equation,
z 2 p2 + z 2 q 2 + z 2 = 1
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
4 / 55
Formation of First and Second order equations
Example: Eliminate the arbitrary constants a & b from (x − a)2 + (y − b)2 + z 2 = 1
and form the PDE.
Solution: Differentiating partially w.r.t. x and y,
2(x − a) + 2zp = 0
2(y − b) + 2zq = 0
Substituting the values in the given equation,
z 2 p2 + z 2 q 2 + z 2 = 1
z 2 (p2 + q 2 + 1) = 1
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
4 / 55
Formation of First and Second order equations
Example: Form the partial differential equation by eliminating the arbitrary
function from z = (x + y)φ(x2 − y 2 ).
(P P Savani University)
Partial Differential Equation
March 16, 2022
J
5 / 55
Formation of First and Second order equations
Example: Form the partial differential equation by eliminating the arbitrary
function from z = (x + y)φ(x2 − y 2 ).
Solution: Differentiating z partially w.r.t x and y,
(P P Savani University)
Partial Differential Equation
March 16, 2022
J
5 / 55
Formation of First and Second order equations
Example: Form the partial differential equation by eliminating the arbitrary
function from z = (x + y)φ(x2 − y 2 ).
Solution: Differentiating z partially w.r.t x and y,
∂z
p=
= (x + y)φ0 (x2 − y 2 ) · 2x + φ(x2 − y 2 )
∂x
(P P Savani University)
Partial Differential Equation
March 16, 2022
J
5 / 55
Formation of First and Second order equations
Example: Form the partial differential equation by eliminating the arbitrary
function from z = (x + y)φ(x2 − y 2 ).
Solution: Differentiating z partially w.r.t x and y,
∂z
p=
= (x + y)φ0 (x2 − y 2 ) · 2x + φ(x2 − y 2 )
∂x
∂z
q=
= (x + y)φ0 (x2 − y 2 ) · (−2y) + φ(x2 − y 2 )
∂y
(P P Savani University)
Partial Differential Equation
March 16, 2022
J
5 / 55
Formation of First and Second order equations
Example: Form the partial differential equation by eliminating the arbitrary
function from z = (x + y)φ(x2 − y 2 ).
Solution: Differentiating z partially w.r.t x and y,
∂z
p=
= (x + y)φ0 (x2 − y 2 ) · 2x + φ(x2 − y 2 )
∂x
∂z
q=
= (x + y)φ0 (x2 − y 2 ) · (−2y) + φ(x2 − y 2 )
∂y
from the above equations
(P P Savani University)
Partial Differential Equation
March 16, 2022
J
5 / 55
Formation of First and Second order equations
Example: Form the partial differential equation by eliminating the arbitrary
function from z = (x + y)φ(x2 − y 2 ).
Solution: Differentiating z partially w.r.t x and y,
∂z
p=
= (x + y)φ0 (x2 − y 2 ) · 2x + φ(x2 − y 2 )
∂x
∂z
q=
= (x + y)φ0 (x2 − y 2 ) · (−2y) + φ(x2 − y 2 )
∂y
z
from the above equations p −
= 2x(x + y)φ0 (x2 − y 2 )
x+y
(P P Savani University)
Partial Differential Equation
March 16, 2022
J
5 / 55
Formation of First and Second order equations
Example: Form the partial differential equation by eliminating the arbitrary
function from z = (x + y)φ(x2 − y 2 ).
Solution: Differentiating z partially w.r.t x and y,
∂z
p=
= (x + y)φ0 (x2 − y 2 ) · 2x + φ(x2 − y 2 )
∂x
∂z
q=
= (x + y)φ0 (x2 − y 2 ) · (−2y) + φ(x2 − y 2 )
∂y
z
from the above equations p −
= 2x(x + y)φ0 (x2 − y 2 )
x+y
z
q−
= −2y(x + y)φ0 (x2 − y 2 )
x+y
(P P Savani University)
Partial Differential Equation
March 16, 2022
J
5 / 55
Formation of First and Second order equations
Example: Form the partial differential equation by eliminating the arbitrary
function from z = (x + y)φ(x2 − y 2 ).
Solution: Differentiating z partially w.r.t x and y,
∂z
p=
= (x + y)φ0 (x2 − y 2 ) · 2x + φ(x2 − y 2 )
∂x
∂z
q=
= (x + y)φ0 (x2 − y 2 ) · (−2y) + φ(x2 − y 2 )
∂y
z
from the above equations p −
= 2x(x + y)φ0 (x2 − y 2 )
x+y
z
q−
= −2y(x + y)φ0 (x2 − y 2 )
x+y
p − z/(x + y)
x
Division gives
=−
q − z/(x + y)
y
(P P Savani University)
Partial Differential Equation
March 16, 2022
J
5 / 55
Formation of First and Second order equations
Example: Form the partial differential equation by eliminating the arbitrary
function from z = (x + y)φ(x2 − y 2 ).
Solution: Differentiating z partially w.r.t x and y,
∂z
p=
= (x + y)φ0 (x2 − y 2 ) · 2x + φ(x2 − y 2 )
∂x
∂z
q=
= (x + y)φ0 (x2 − y 2 ) · (−2y) + φ(x2 − y 2 )
∂y
z
from the above equations p −
= 2x(x + y)φ0 (x2 − y 2 )
x+y
z
q−
= −2y(x + y)φ0 (x2 − y 2 )
x+y
p − z/(x + y)
x
Division gives
=−
i.e., [p(x + y) − z]y + [q(x + y) − z]x = 0
q − z/(x + y)
y
(P P Savani University)
Partial Differential Equation
March 16, 2022
J
5 / 55
Formation of First and Second order equations
Example: Form the partial differential equation by eliminating the arbitrary
function from z = (x + y)φ(x2 − y 2 ).
Solution: Differentiating z partially w.r.t x and y,
∂z
p=
= (x + y)φ0 (x2 − y 2 ) · 2x + φ(x2 − y 2 )
∂x
∂z
q=
= (x + y)φ0 (x2 − y 2 ) · (−2y) + φ(x2 − y 2 )
∂y
z
from the above equations p −
= 2x(x + y)φ0 (x2 − y 2 )
x+y
z
q−
= −2y(x + y)φ0 (x2 − y 2 )
x+y
p − z/(x + y)
x
Division gives
=−
i.e., [p(x + y) − z]y + [q(x + y) − z]x = 0
q − z/(x + y)
y
(x + y)(py + qx) − z(x + y) = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
J
5 / 55
Formation of First and Second order equations
Example: Form the partial differential equation by eliminating the arbitrary
function from z = (x + y)φ(x2 − y 2 ).
Solution: Differentiating z partially w.r.t x and y,
∂z
p=
= (x + y)φ0 (x2 − y 2 ) · 2x + φ(x2 − y 2 )
∂x
∂z
q=
= (x + y)φ0 (x2 − y 2 ) · (−2y) + φ(x2 − y 2 )
∂y
z
from the above equations p −
= 2x(x + y)φ0 (x2 − y 2 )
x+y
z
q−
= −2y(x + y)φ0 (x2 − y 2 )
x+y
p − z/(x + y)
x
Division gives
=−
i.e., [p(x + y) − z]y + [q(x + y) − z]x = 0
q − z/(x + y)
y
(x + y)(py + qx) − z(x + y) = 0
(P P Savani University)
Hence, py + qx = z is the required equation.
Partial Differential Equation
March 16, 2022
J
5 / 55
Solution of Partial Differential Equations
Solution of PDE
The solution of a partial differential equation is a relation between the dependent
and independent variables, which satisfies the equation.
The solution of a partial differential equation is not always unique. It may have
more than one solution or sometimes no solution.
(P P Savani University)
Partial Differential Equation
March 16, 2022
6 / 55
Solution of Partial Differential Equations
Solution of PDE
The solution of a partial differential equation is a relation between the dependent
and independent variables, which satisfies the equation.
The solution of a partial differential equation is not always unique. It may have
more than one solution or sometimes no solution.
Complete Integral:
The solution f (x, y, z, a, b) = 0 of a first order partial differential equation
which contains two arbitrary constants.
Note: No. of arbitrary constants equal to the independent variables.
(P P Savani University)
Partial Differential Equation
March 16, 2022
6 / 55
Solution of Partial Differential Equations
Particular Integral:
A solution obtained from the complete integral by assigning particular values
to the arbitrary constants.
(P P Savani University)
Partial Differential Equation
March 16, 2022
7 / 55
Solution of Partial Differential Equations
Particular Integral:
A solution obtained from the complete integral by assigning particular values
to the arbitrary constants.
General Integral:
A solution, containing an arbitrary function φ, obtained by putting b = φ(a)
in above equation and then finding f [x, y, z, φ(a)] = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
7 / 55
Solution of Partial Differential Equations
Solution by Direct Integration
Example: Solve
∂ 3z
+ 18xy 2 + sin(2x − y) = 0
∂ 2 x∂y
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
8 / 55
Solution of Partial Differential Equations
Solution by Direct Integration
∂ 3z
+ 18xy 2 + sin(2x − y) = 0
∂ 2 x∂y
Solution: Integrating twice w.r.t. x (keeping y fixed),
Example: Solve
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
8 / 55
Solution of Partial Differential Equations
Solution by Direct Integration
∂ 3z
+ 18xy 2 + sin(2x − y) = 0
∂ 2 x∂y
Solution: Integrating twice w.r.t. x (keeping y fixed),
Example: Solve
∂ 2z
1
+ 9x2 y 2 − cos(2x − y) = f (y)
∂x∂y
2
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
8 / 55
Solution of Partial Differential Equations
Solution by Direct Integration
∂ 3z
+ 18xy 2 + sin(2x − y) = 0
∂ 2 x∂y
Solution: Integrating twice w.r.t. x (keeping y fixed),
Example: Solve
∂ 2z
1
+ 9x2 y 2 − cos(2x − y) = f (y)
∂x∂y
2
∂z
1
3 2
+ 3x y − sin(2x − y) = xf (y) + g(y)
∂y
4
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
8 / 55
Solution of Partial Differential Equations
Solution by Direct Integration
∂ 3z
+ 18xy 2 + sin(2x − y) = 0
∂ 2 x∂y
Solution: Integrating twice w.r.t. x (keeping y fixed),
Example: Solve
∂ 2z
1
+ 9x2 y 2 − cos(2x − y) = f (y)
∂x∂y
2
∂z
1
3 2
+ 3x y − sin(2x − y) = xf (y) + g(y)
∂y
4
Now integrating w.r.t y (keeping x fixed),
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
8 / 55
Solution of Partial Differential Equations
Solution by Direct Integration
∂ 3z
+ 18xy 2 + sin(2x − y) = 0
∂ 2 x∂y
Solution: Integrating twice w.r.t. x (keeping y fixed),
Example: Solve
∂ 2z
1
+ 9x2 y 2 − cos(2x − y) = f (y)
∂x∂y
2
∂z
1
3 2
+ 3x y − sin(2x − y) = xf (y) + g(y)
∂y
4
Now integrating w.r.t y (keeping x fixed),
Z
Z
1
z + x3 y 3 − cos(2x − y) = x f (y)dy + g(y)dy + w(x)
4
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
8 / 55
Solution of Partial Differential Equations
Solution by Direct Integration
∂ 3z
+ 18xy 2 + sin(2x − y) = 0
∂ 2 x∂y
Solution: Integrating twice w.r.t. x (keeping y fixed),
Example: Solve
∂ 2z
1
+ 9x2 y 2 − cos(2x − y) = f (y)
∂x∂y
2
∂z
1
3 2
+ 3x y − sin(2x − y) = xf (y) + g(y)
∂y
4
Now integrating w.r.t y (keeping x fixed),
Z
Z
1
z + x3 y 3 − cos(2x − y) = x f (y)dy + g(y)dy + w(x)
4
Z
Z
Let
f (y)dy = u(y) and
g(y)dy = v(y)
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
8 / 55
Solution of Partial Differential Equations
Solution by Direct Integration
∂ 3z
+ 18xy 2 + sin(2x − y) = 0
∂ 2 x∂y
Solution: Integrating twice w.r.t. x (keeping y fixed),
Example: Solve
∂ 2z
1
+ 9x2 y 2 − cos(2x − y) = f (y)
∂x∂y
2
∂z
1
3 2
+ 3x y − sin(2x − y) = xf (y) + g(y)
∂y
4
Now integrating w.r.t y (keeping x fixed),
Z
Z
1
z + x3 y 3 − cos(2x − y) = x f (y)dy + g(y)dy + w(x)
4
Z
Z
Let
f (y)dy = u(y) and
g(y)dy = v(y)
Hence, z =
1
cos(2x − y) − x3 y 3 + xu(y) + v(y) + w(x) where u, v, w are arbitrary functions.
4
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
8 / 55
Solution of Partial Differential Equations
Examples for Practice
Solve the following PDEs:
∂ 2z
x
1
= +a
∂x∂y
y
2
∂ z
2
= xy
∂y∂x
∂ 2u
3
= e−t cos x
∂x∂t
(P P Savani University)
Partial Differential Equation
March 16, 2022
9 / 55
Linear PDEs
Linear Partial Differential Equations of First Order
A PDE of first order is said to be linear if the dependent variable and its derivatives
are of degree one and the products of the dependent variable and its derivatives do
not appear in the equation.
A Linear PDE of the first order,
P p + Qq = R
where P , Q and R are functions of x, y, z. This equation is called quasi-linear equation
as the degree of the highest-order derivative is one and the products of the highest-order
partial derivatives are not present. When P, Q and R are independent of z it is known as
Lagrange’s linear equation.
The general solution of Lagrange’s linear equation is given by f (u, v) = 0 or u = φ(v) where
f is an arbitrary function and u, v are functions of x, y and z.
(P P Savani University)
Partial Differential Equation
March 16, 2022
10 / 55
Linear PDEs
How to solve?
Working Rules for Solving Lagrange’s Linear Equations
1
Write the given differential equation in the standard form P p + Qq = R.
2
Form the Lagrange’s auxiliary equation (subsidiary equation)
dx
dy
dz
=
=
P
Q
R
3
Solve the auxiliary equation to obtain its two independent solutions as
u = c1 , v = c2
4
Write the general solution of the given equation as
f (u, v) = 0 or u = φ(v)
(P P Savani University)
Partial Differential Equation
March 16, 2022
11 / 55
Linear PDEs
dx
dy
dz
=
=
P
Q
R
To solve above type of problems we have following methods:
(1) Method of Grouping:
In some problems, it is possible to solve any two of these equations
Solution of the equation of the type
dx
dy
=
P
Q
or
dy
dz
=
Q
R
or
dx
dz
=
P
R
In such cases, solve any of the differential equation, get the solution and then substitute
in the other differential equation.
dz
For eg., See if it is possible to take two fractions dx
P = R from which y can be cancelled or
is absent, leaving equations in x and z only. If so, obtain the first solution φ(x, z) = c1 , by
integrating the fractions.
Similarly, take any other two fractions and obtain the second solution ψ(y, z) = c2 .
These two independent solutions taken together constitute the complete solution required.
(P P Savani University)
Partial Differential Equation
March 16, 2022
12 / 55
Linear PDEs
(2) Method of Multiplier:
By a proper choice of the multipliers l, m, n which are not necessarily constants, we
can write
dx
dy
dz
ldx + mdy + ndz
=
=
=
P
Q
R
lP + mQ + nR
such that lP + mQ + nR = 0.
Then obtain the first solution φ(x, y, z) = c1 , by integrating ldx + mdy + ndz = 0.
Similarly, choosing another set of multipliers λ, µ, γ and obtain the second solution
ψ(x, y, z) = c2
These two independent solutions taken together constitute the complete solution required.
(P P Savani University)
Partial Differential Equation
March 16, 2022
13 / 55
Linear PDEs
2
Example: Solve
y z
p + xzq = y 2
x
(P P Savani University)
Partial Differential Equation
March 16, 2022
14 / 55
Linear PDEs
2
y z
p + xzq = y 2
x
Solution: Rewriting the equation as y 2 zp + x2 zq = y 2 x
Example: Solve
(P P Savani University)
Partial Differential Equation
March 16, 2022
14 / 55
Linear PDEs
2
y z
p + xzq = y 2
x
Solution: Rewriting the equation as y 2 zp + x2 zq = y 2 x
Example: Solve
P = y 2 z,
(P P Savani University)
Q = x2 z,
R = y2x
Partial Differential Equation
March 16, 2022
14 / 55
Linear PDEs
2
y z
p + xzq = y 2
x
Solution: Rewriting the equation as y 2 zp + x2 zq = y 2 x
Example: Solve
P = y 2 z,
Q = x2 z,
R = y2x
The subsidiary equations are
dx
dy
dz
= 2 = 2
2
y z
xz
y x
(P P Savani University)
Partial Differential Equation
March 16, 2022
14 / 55
Linear PDEs
2
y z
p + xzq = y 2
x
Solution: Rewriting the equation as y 2 zp + x2 zq = y 2 x
Example: Solve
P = y 2 z,
Q = x2 z,
R = y2x
The subsidiary equations are
dx
dy
dz
= 2 = 2
2
y z
xz
y x
Taking the first and second fractions,
(P P Savani University)
Partial Differential Equation
March 16, 2022
14 / 55
Linear PDEs
2
y z
p + xzq = y 2
x
Solution: Rewriting the equation as y 2 zp + x2 zq = y 2 x
Example: Solve
P = y 2 z,
Q = x2 z,
R = y2x
The subsidiary equations are
dx
dy
dz
= 2 = 2
2
y z
xz
y x
Taking the first and second fractions,
dx
dy
=
y2z
x2 z
(P P Savani University)
Partial Differential Equation
March 16, 2022
14 / 55
Linear PDEs
2
y z
p + xzq = y 2
x
Solution: Rewriting the equation as y 2 zp + x2 zq = y 2 x
Example: Solve
P = y 2 z,
Q = x2 z,
R = y2x
The subsidiary equations are
dx
dy
dz
= 2 = 2
2
y z
xz
y x
Taking the first and second fractions,
dx
dy
=
y2z
x2 z
x2 dx = y 2 dy
(P P Savani University)
Partial Differential Equation
March 16, 2022
14 / 55
Linear PDEs
Integrating,
x3 − y 3 = a
(P P Savani University)
Partial Differential Equation
March 16, 2022
15 / 55
Linear PDEs
Integrating,
x3 − y 3 = a
Taking the first and third fractions,
(P P Savani University)
Partial Differential Equation
March 16, 2022
15 / 55
Linear PDEs
Integrating,
x3 − y 3 = a
Taking the first and third fractions,
dx
dz
= 2
2
y z
y x
(P P Savani University)
Partial Differential Equation
March 16, 2022
15 / 55
Linear PDEs
Integrating,
x3 − y 3 = a
Taking the first and third fractions,
dx
dz
= 2
2
y z
y x
xdx = zdz
(P P Savani University)
Partial Differential Equation
March 16, 2022
15 / 55
Linear PDEs
Integrating,
x3 − y 3 = a
Taking the first and third fractions,
dx
dz
= 2
2
y z
y x
xdx = zdz
Integrating,
(P P Savani University)
Partial Differential Equation
March 16, 2022
15 / 55
Linear PDEs
Integrating,
x3 − y 3 = a
Taking the first and third fractions,
dx
dz
= 2
2
y z
y x
xdx = zdz
Integrating,
x2 − z 2 = b
(P P Savani University)
Partial Differential Equation
March 16, 2022
15 / 55
Linear PDEs
Integrating,
x3 − y 3 = a
Taking the first and third fractions,
dx
dz
= 2
2
y z
y x
xdx = zdz
Integrating,
x2 − z 2 = b
Hence, the general solution is
(P P Savani University)
Partial Differential Equation
March 16, 2022
15 / 55
Linear PDEs
Integrating,
x3 − y 3 = a
Taking the first and third fractions,
dx
dz
= 2
2
y z
y x
xdx = zdz
Integrating,
x2 − z 2 = b
Hence, the general solution is
x3 − y 3 = φ(x2 − z 2 )
or
f (x3 − y 3 , x2 − z 2 ) = 0
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
15 / 55
Linear PDEs
Example: Solve yzp − xzq = xy
(P P Savani University)
Partial Differential Equation
March 16, 2022
16 / 55
Linear PDEs
Example: Solve yzp − xzq = xy
Solution: The subsidiary equations are
(P P Savani University)
Partial Differential Equation
March 16, 2022
16 / 55
Linear PDEs
Example: Solve yzp − xzq = xy
Solution: The subsidiary equations are
dx
dy
dz
=
=
yz
−xz
xy
(P P Savani University)
Partial Differential Equation
March 16, 2022
16 / 55
Linear PDEs
Example: Solve yzp − xzq = xy
Solution: The subsidiary equations are
dx
dy
dz
=
=
yz
−xz
xy
Taking the first and second fractions,
(P P Savani University)
Partial Differential Equation
March 16, 2022
16 / 55
Linear PDEs
Example: Solve yzp − xzq = xy
Solution: The subsidiary equations are
dx
dy
dz
=
=
yz
−xz
xy
Taking the first and second fractions,
dx
dy
=
yz
−xz
(P P Savani University)
Partial Differential Equation
March 16, 2022
16 / 55
Linear PDEs
Example: Solve yzp − xzq = xy
Solution: The subsidiary equations are
dx
dy
dz
=
=
yz
−xz
xy
Taking the first and second fractions,
dx
dy
=
yz
−xz
xdx + ydy = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
16 / 55
Linear PDEs
Example: Solve yzp − xzq = xy
Solution: The subsidiary equations are
dx
dy
dz
=
=
yz
−xz
xy
Taking the first and second fractions,
dx
dy
=
yz
−xz
xdx + ydy = 0
Integrating,
(P P Savani University)
Partial Differential Equation
March 16, 2022
16 / 55
Linear PDEs
Example: Solve yzp − xzq = xy
Solution: The subsidiary equations are
dx
dy
dz
=
=
yz
−xz
xy
Taking the first and second fractions,
dx
dy
=
yz
−xz
xdx + ydy = 0
Integrating,
x2 y 2
+
=c
2
2
(P P Savani University)
Partial Differential Equation
March 16, 2022
16 / 55
Linear PDEs
Example: Solve yzp − xzq = xy
Solution: The subsidiary equations are
dx
dy
dz
=
=
yz
−xz
xy
Taking the first and second fractions,
dx
dy
=
yz
−xz
xdx + ydy = 0
Integrating,
x2 y 2
+
=c
2
2
(P P Savani University)
i.e.,
x 2 + y 2 = c1
Partial Differential Equation
(where c1 = 2c)
March 16, 2022
16 / 55
Linear PDEs
Taking the second and third fractions,
(P P Savani University)
Partial Differential Equation
March 16, 2022
17 / 55
Linear PDEs
Taking the second and third fractions,
dz
dy
=
−xz
xy
(P P Savani University)
Partial Differential Equation
March 16, 2022
17 / 55
Linear PDEs
Taking the second and third fractions,
dz
dy
=
−xz
xy
ydy + zdz = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
17 / 55
Linear PDEs
Taking the second and third fractions,
dz
dy
=
−xz
xy
ydy + zdz = 0
Integrating,
(P P Savani University)
Partial Differential Equation
March 16, 2022
17 / 55
Linear PDEs
Taking the second and third fractions,
dz
dy
=
−xz
xy
ydy + zdz = 0
Integrating,
y2 z2
+
= c0
2
2
(P P Savani University)
Partial Differential Equation
March 16, 2022
17 / 55
Linear PDEs
Taking the second and third fractions,
dz
dy
=
−xz
xy
ydy + zdz = 0
Integrating,
y2 z2
+
= c0
2
2
y 2 + z 2 = c2
(P P Savani University)
(where c2 = 2c0 )
Partial Differential Equation
March 16, 2022
17 / 55
Linear PDEs
Taking the second and third fractions,
dz
dy
=
−xz
xy
ydy + zdz = 0
Integrating,
y2 z2
+
= c0
2
2
y 2 + z 2 = c2
(where c2 = 2c0 )
Hence, the general solution is
(P P Savani University)
Partial Differential Equation
March 16, 2022
17 / 55
Linear PDEs
Taking the second and third fractions,
dz
dy
=
−xz
xy
ydy + zdz = 0
Integrating,
y2 z2
+
= c0
2
2
y 2 + z 2 = c2
(where c2 = 2c0 )
Hence, the general solution is
x2 + y 2 = φ(y 2 + z 2 )
or
f (x2 + y 2 , y 2 + z 2 ) = 0
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
17 / 55
Linear PDEs
Example: Solve (mz − ny)
(P P Savani University)
∂z
∂z
+ (nx − lz)
= ly − mx
∂x
∂y
Partial Differential Equation
March 16, 2022
18 / 55
Linear PDEs
∂z
∂z
+ (nx − lz)
= ly − mx
∂x
∂y
Solution: The subsidiary equations are
Example: Solve (mz − ny)
(P P Savani University)
Partial Differential Equation
March 16, 2022
18 / 55
Linear PDEs
∂z
∂z
+ (nx − lz)
= ly − mx
∂x
∂y
Solution: The subsidiary equations are
Example: Solve (mz − ny)
dx
dy
dz
=
=
mz − ny
nx − lz
ly − mx
(P P Savani University)
Partial Differential Equation
March 16, 2022
18 / 55
Linear PDEs
∂z
∂z
+ (nx − lz)
= ly − mx
∂x
∂y
Solution: The subsidiary equations are
Example: Solve (mz − ny)
dx
dy
dz
=
=
mz − ny
nx − lz
ly − mx
Taking x, y and z as multipliers,
(P P Savani University)
Partial Differential Equation
March 16, 2022
18 / 55
Linear PDEs
∂z
∂z
+ (nx − lz)
= ly − mx
∂x
∂y
Solution: The subsidiary equations are
Example: Solve (mz − ny)
dx
dy
dz
=
=
mz − ny
nx − lz
ly − mx
Taking x, y and z as multipliers,
dx
dy
dz
xdx + ydy + zdz
=
=
=
mz − ny
nx − lz
ly − mx
0
(P P Savani University)
Partial Differential Equation
March 16, 2022
18 / 55
Linear PDEs
∂z
∂z
+ (nx − lz)
= ly − mx
∂x
∂y
Solution: The subsidiary equations are
Example: Solve (mz − ny)
dx
dy
dz
=
=
mz − ny
nx − lz
ly − mx
Taking x, y and z as multipliers,
dx
dy
dz
xdx + ydy + zdz
=
=
=
mz − ny
nx − lz
ly − mx
0
∴ xdx + ydy + zdz = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
18 / 55
Linear PDEs
∂z
∂z
+ (nx − lz)
= ly − mx
∂x
∂y
Solution: The subsidiary equations are
Example: Solve (mz − ny)
dx
dy
dz
=
=
mz − ny
nx − lz
ly − mx
Taking x, y and z as multipliers,
dx
dy
dz
xdx + ydy + zdz
=
=
=
mz − ny
nx − lz
ly − mx
0
∴ xdx + ydy + zdz = 0
Integrating,
x2 + y 2 + z 2 = a
(P P Savani University)
Partial Differential Equation
March 16, 2022
18 / 55
Linear PDEs
Taking l, m and n as multipliers,
(P P Savani University)
Partial Differential Equation
March 16, 2022
19 / 55
Linear PDEs
Taking l, m and n as multipliers,
dx
dy
dz
ldx + mdy + ndz
=
=
=
mz − ny
nx − lz
ly − mx
0
(P P Savani University)
Partial Differential Equation
March 16, 2022
19 / 55
Linear PDEs
Taking l, m and n as multipliers,
dx
dy
dz
ldx + mdy + ndz
=
=
=
mz − ny
nx − lz
ly − mx
0
∴ ldx + mdy + ndz = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
19 / 55
Linear PDEs
Taking l, m and n as multipliers,
dx
dy
dz
ldx + mdy + ndz
=
=
=
mz − ny
nx − lz
ly − mx
0
∴ ldx + mdy + ndz = 0
Integrating,
(P P Savani University)
Partial Differential Equation
March 16, 2022
19 / 55
Linear PDEs
Taking l, m and n as multipliers,
dx
dy
dz
ldx + mdy + ndz
=
=
=
mz − ny
nx − lz
ly − mx
0
∴ ldx + mdy + ndz = 0
Integrating,
lx + my + nz = b
(P P Savani University)
Partial Differential Equation
March 16, 2022
19 / 55
Linear PDEs
Taking l, m and n as multipliers,
dx
dy
dz
ldx + mdy + ndz
=
=
=
mz − ny
nx − lz
ly − mx
0
∴ ldx + mdy + ndz = 0
Integrating,
lx + my + nz = b
Hence, the general solution is
(P P Savani University)
Partial Differential Equation
March 16, 2022
19 / 55
Linear PDEs
Taking l, m and n as multipliers,
dx
dy
dz
ldx + mdy + ndz
=
=
=
mz − ny
nx − lz
ly − mx
0
∴ ldx + mdy + ndz = 0
Integrating,
lx + my + nz = b
Hence, the general solution is
x2 + y 2 + z 2 = φ(lx + my + nz)
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
19 / 55
Linear PDEs
Example: Solve (x2 − y 2 − z 2 )p + 2xyq = 2xz
(P P Savani University)
Partial Differential Equation
March 16, 2022
20 / 55
Linear PDEs
Example: Solve (x2 − y 2 − z 2 )p + 2xyq = 2xz
Solution: The subsidiary equations are
(P P Savani University)
Partial Differential Equation
March 16, 2022
20 / 55
Linear PDEs
Example: Solve (x2 − y 2 − z 2 )p + 2xyq = 2xz
Solution: The subsidiary equations are
dx
dy
dz
=
=
2
2
2
x −y −z
2xy
2xz
(P P Savani University)
Partial Differential Equation
March 16, 2022
20 / 55
Linear PDEs
Example: Solve (x2 − y 2 − z 2 )p + 2xyq = 2xz
Solution: The subsidiary equations are
dx
dy
dz
=
=
2
2
2
x −y −z
2xy
2xz
Taking x, y and z as multipliers,
(P P Savani University)
Partial Differential Equation
March 16, 2022
20 / 55
Linear PDEs
Example: Solve (x2 − y 2 − z 2 )p + 2xyq = 2xz
Solution: The subsidiary equations are
dx
dy
dz
=
=
2
2
2
x −y −z
2xy
2xz
Taking x, y and z as multipliers,
dx
dy
dz
xdx + ydy + zdz
xdx + ydy + zdz
=
=
=
=
x2 − y 2 − z 2
2xy
2xz
x3 − xy 2 − xz 2 + 2xy 2 + 2xz 2
x3 + xy 2 + xz 2
(P P Savani University)
Partial Differential Equation
March 16, 2022
20 / 55
Linear PDEs
Example: Solve (x2 − y 2 − z 2 )p + 2xyq = 2xz
Solution: The subsidiary equations are
dx
dy
dz
=
=
2
2
2
x −y −z
2xy
2xz
Taking x, y and z as multipliers,
dx
dy
dz
xdx + ydy + zdz
xdx + ydy + zdz
=
=
=
=
x2 − y 2 − z 2
2xy
2xz
x3 − xy 2 − xz 2 + 2xy 2 + 2xz 2
x3 + xy 2 + xz 2
Taking the second and third fractions,
(P P Savani University)
Partial Differential Equation
March 16, 2022
20 / 55
Linear PDEs
Example: Solve (x2 − y 2 − z 2 )p + 2xyq = 2xz
Solution: The subsidiary equations are
dx
dy
dz
=
=
2
2
2
x −y −z
2xy
2xz
Taking x, y and z as multipliers,
dx
dy
dz
xdx + ydy + zdz
xdx + ydy + zdz
=
=
=
=
x2 − y 2 − z 2
2xy
2xz
x3 − xy 2 − xz 2 + 2xy 2 + 2xz 2
x3 + xy 2 + xz 2
Taking the second and third fractions,
dy
dz
dy
dz
=
i.e.,
=
2xy
2xz
y
z
(P P Savani University)
Partial Differential Equation
March 16, 2022
20 / 55
Linear PDEs
Example: Solve (x2 − y 2 − z 2 )p + 2xyq = 2xz
Solution: The subsidiary equations are
dx
dy
dz
=
=
2
2
2
x −y −z
2xy
2xz
Taking x, y and z as multipliers,
dx
dy
dz
xdx + ydy + zdz
xdx + ydy + zdz
=
=
=
=
x2 − y 2 − z 2
2xy
2xz
x3 − xy 2 − xz 2 + 2xy 2 + 2xz 2
x3 + xy 2 + xz 2
Taking the second and third fractions,
dy
dz
dy
dz
=
i.e.,
=
2xy
2xz
y
z
Integrating,
(P P Savani University)
Partial Differential Equation
March 16, 2022
20 / 55
Linear PDEs
Example: Solve (x2 − y 2 − z 2 )p + 2xyq = 2xz
Solution: The subsidiary equations are
dx
dy
dz
=
=
2
2
2
x −y −z
2xy
2xz
Taking x, y and z as multipliers,
dx
dy
dz
xdx + ydy + zdz
xdx + ydy + zdz
=
=
=
=
x2 − y 2 − z 2
2xy
2xz
x3 − xy 2 − xz 2 + 2xy 2 + 2xz 2
x3 + xy 2 + xz 2
Taking the second and third fractions,
dy
dz
dy
dz
=
i.e.,
=
2xy
2xz
y
z
Integrating,
log y = log z + log c1
(P P Savani University)
Partial Differential Equation
March 16, 2022
20 / 55
Linear PDEs
Example: Solve (x2 − y 2 − z 2 )p + 2xyq = 2xz
Solution: The subsidiary equations are
dx
dy
dz
=
=
2
2
2
x −y −z
2xy
2xz
Taking x, y and z as multipliers,
dx
dy
dz
xdx + ydy + zdz
xdx + ydy + zdz
=
=
=
=
x2 − y 2 − z 2
2xy
2xz
x3 − xy 2 − xz 2 + 2xy 2 + 2xz 2
x3 + xy 2 + xz 2
Taking the second and third fractions,
dy
dz
dy
dz
=
i.e.,
=
2xy
2xz
y
z
Integrating,
log y = log z + log c1
y
log = log c1
z
(P P Savani University)
Partial Differential Equation
March 16, 2022
20 / 55
Linear PDEs
Example: Solve (x2 − y 2 − z 2 )p + 2xyq = 2xz
Solution: The subsidiary equations are
dx
dy
dz
=
=
2
2
2
x −y −z
2xy
2xz
Taking x, y and z as multipliers,
dx
dy
dz
xdx + ydy + zdz
xdx + ydy + zdz
=
=
=
=
x2 − y 2 − z 2
2xy
2xz
x3 − xy 2 − xz 2 + 2xy 2 + 2xz 2
x3 + xy 2 + xz 2
Taking the second and third fractions,
dy
dz
dy
dz
=
i.e.,
=
2xy
2xz
y
z
Integrating,
log y = log z + log c1
y
y
log = log c1
i.e.,
= c1
z
z
(P P Savani University)
Partial Differential Equation
March 16, 2022
20 / 55
Linear PDEs
Taking the third and fifth fractions,
(P P Savani University)
Partial Differential Equation
March 16, 2022
21 / 55
Linear PDEs
Taking the third and fifth fractions,
dz
xdx + ydy + zdz
= 3
2xz
x + xy 2 + xz 2
(P P Savani University)
Partial Differential Equation
March 16, 2022
21 / 55
Linear PDEs
Taking the third and fifth fractions,
dz
xdx + ydy + zdz
= 3
2xz
x + xy 2 + xz 2
dz
xdx + ydy + zdz
=
2xz
x(x2 + y 2 + z 2 )
(P P Savani University)
Partial Differential Equation
March 16, 2022
21 / 55
Linear PDEs
Taking the third and fifth fractions,
dz
xdx + ydy + zdz
= 3
2xz
x + xy 2 + xz 2
dz
xdx + ydy + zdz
=
2xz
x(x2 + y 2 + z 2 )
dz
2xdx + 2ydy + 2zdz
=
z
x2 + y 2 + z 2
(P P Savani University)
Partial Differential Equation
March 16, 2022
21 / 55
Linear PDEs
Taking the third and fifth fractions,
dz
xdx + ydy + zdz
= 3
2xz
x + xy 2 + xz 2
dz
xdx + ydy + zdz
=
2xz
x(x2 + y 2 + z 2 )
dz
2xdx + 2ydy + 2zdz
=
z
x2 + y 2 + z 2
Integrating,
(P P Savani University)
Partial Differential Equation
March 16, 2022
21 / 55
Linear PDEs
Taking the third and fifth fractions,
dz
xdx + ydy + zdz
= 3
2xz
x + xy 2 + xz 2
dz
xdx + ydy + zdz
=
2xz
x(x2 + y 2 + z 2 )
dz
2xdx + 2ydy + 2zdz
=
z
x2 + y 2 + z 2
Integrating,
log z = log(x2 + y 2 + z 2 ) + log c2
(P P Savani University)
Partial Differential Equation
March 16, 2022
21 / 55
Linear PDEs
Taking the third and fifth fractions,
dz
xdx + ydy + zdz
= 3
2xz
x + xy 2 + xz 2
dz
xdx + ydy + zdz
=
2xz
x(x2 + y 2 + z 2 )
dz
2xdx + 2ydy + 2zdz
=
z
x2 + y 2 + z 2
Integrating,
log z = log(x2 + y 2 + z 2 ) + log c2
z
log
= log c2
x2 + y 2 + z 2
(P P Savani University)
Partial Differential Equation
March 16, 2022
21 / 55
Linear PDEs
Taking the third and fifth fractions,
dz
xdx + ydy + zdz
= 3
2xz
x + xy 2 + xz 2
dz
xdx + ydy + zdz
=
2xz
x(x2 + y 2 + z 2 )
dz
2xdx + 2ydy + 2zdz
=
z
x2 + y 2 + z 2
Integrating,
log z = log(x2 + y 2 + z 2 ) + log c2
z
z
log
= log c2 ⇒
= c2
2
2
2
2
x +y +z
x + y2 + z2
(P P Savani University)
Partial Differential Equation
March 16, 2022
21 / 55
Linear PDEs
Taking the third and fifth fractions,
dz
xdx + ydy + zdz
= 3
2xz
x + xy 2 + xz 2
dz
xdx + ydy + zdz
=
2xz
x(x2 + y 2 + z 2 )
dz
2xdx + 2ydy + 2zdz
=
z
x2 + y 2 + z 2
Integrating,
log z = log(x2 + y 2 + z 2 ) + log c2
z
z
log
= log c2 ⇒
= c2
2
2
2
2
x +y +z
x + y2 + z2
Hence, the general solution is
(P P Savani University)
Partial Differential Equation
March 16, 2022
21 / 55
Linear PDEs
Taking the third and fifth fractions,
dz
xdx + ydy + zdz
= 3
2xz
x + xy 2 + xz 2
dz
xdx + ydy + zdz
=
2xz
x(x2 + y 2 + z 2 )
dz
2xdx + 2ydy + 2zdz
=
z
x2 + y 2 + z 2
Integrating,
log z = log(x2 + y 2 + z 2 ) + log c2
z
z
log
= log c2 ⇒
= c2
2
2
2
2
x +y +z
x + y2 + z2
Hence, the general solution is y
z
f
,
=0
z x2 + y 2 + z 2
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
21 / 55
Linear PDEs
Example: Solve x2 (y − z)p + y 2 (z − x)q = z 2 (x − y)
(P P Savani University)
Partial Differential Equation
March 16, 2022
22 / 55
Linear PDEs
Example: Solve x2 (y − z)p + y 2 (z − x)q = z 2 (x − y)
Solution: The subsidiary equations are
(P P Savani University)
Partial Differential Equation
March 16, 2022
22 / 55
Linear PDEs
Example: Solve x2 (y − z)p + y 2 (z − x)q = z 2 (x − y)
Solution: The subsidiary equations are
dx
dy
dz
= 2
= 2
− z)
y (z − x)
z (x − y)
x2 (y
(P P Savani University)
Partial Differential Equation
March 16, 2022
22 / 55
Linear PDEs
Example: Solve x2 (y − z)p + y 2 (z − x)q = z 2 (x − y)
Solution: The subsidiary equations are
dx
dy
dz
= 2
= 2
− z)
y (z − x)
z (x − y)
x2 (y
Taking 1/x, 1/y and 1/z as multipliers,
(P P Savani University)
Partial Differential Equation
March 16, 2022
22 / 55
Linear PDEs
Example: Solve x2 (y − z)p + y 2 (z − x)q = z 2 (x − y)
Solution: The subsidiary equations are
dx
dy
dz
= 2
= 2
− z)
y (z − x)
z (x − y)
x2 (y
Taking 1/x, 1/y and 1/z as multipliers,
1
1
1
dx + dy + dz
dx
dy
dz
x
y
z
= 2
= 2
=
2
x (y − z)
y (z − x)
z (x − y)
0
(P P Savani University)
Partial Differential Equation
March 16, 2022
22 / 55
Linear PDEs
Example: Solve x2 (y − z)p + y 2 (z − x)q = z 2 (x − y)
Solution: The subsidiary equations are
dx
dy
dz
= 2
= 2
− z)
y (z − x)
z (x − y)
x2 (y
Taking 1/x, 1/y and 1/z as multipliers,
1
1
1
dx + dy + dz
dx
dy
dz
x
y
z
= 2
= 2
=
2
x (y − z)
y (z − x)
z (x − y)
0
∴
1
1
1
dx + dy + dz = 0
x
y
z
(P P Savani University)
Partial Differential Equation
March 16, 2022
22 / 55
Linear PDEs
Example: Solve x2 (y − z)p + y 2 (z − x)q = z 2 (x − y)
Solution: The subsidiary equations are
dx
dy
dz
= 2
= 2
− z)
y (z − x)
z (x − y)
x2 (y
Taking 1/x, 1/y and 1/z as multipliers,
1
1
1
dx + dy + dz
dx
dy
dz
x
y
z
= 2
= 2
=
2
x (y − z)
y (z − x)
z (x − y)
0
1
1
1
dx + dy + dz = 0
x
y
z
Integrating,
∴
(P P Savani University)
Partial Differential Equation
March 16, 2022
22 / 55
Linear PDEs
Example: Solve x2 (y − z)p + y 2 (z − x)q = z 2 (x − y)
Solution: The subsidiary equations are
dx
dy
dz
= 2
= 2
− z)
y (z − x)
z (x − y)
x2 (y
Taking 1/x, 1/y and 1/z as multipliers,
1
1
1
dx + dy + dz
dx
dy
dz
x
y
z
= 2
= 2
=
2
x (y − z)
y (z − x)
z (x − y)
0
1
1
1
dx + dy + dz = 0
x
y
z
Integrating,
log x + log y + log z = log a
∴
(P P Savani University)
or
Partial Differential Equation
xyz = a
March 16, 2022
22 / 55
Linear PDEs
Taking 1/x2 , 1/y 2 and 1/z 2 as multipliers,
(P P Savani University)
Partial Differential Equation
March 16, 2022
23 / 55
Linear PDEs
Taking 1/x2 , 1/y 2 and 1/z 2 as multipliers,
1
1
1
dx
+
dy
+
dz
dx
dy
dz
x2
y2
z2
= 2
= 2
=
x2 (y − z)
y (z − x)
z (x − y)
0
(P P Savani University)
Partial Differential Equation
March 16, 2022
23 / 55
Linear PDEs
Taking 1/x2 , 1/y 2 and 1/z 2 as multipliers,
1
1
1
dx
+
dy
+
dz
dx
dy
dz
x2
y2
z2
= 2
= 2
=
x2 (y − z)
y (z − x)
z (x − y)
0
∴
1
1
1
dx + 2 dy + 2 dz = 0
2
x
y
z
(P P Savani University)
Partial Differential Equation
March 16, 2022
23 / 55
Linear PDEs
Taking 1/x2 , 1/y 2 and 1/z 2 as multipliers,
1
1
1
dx
+
dy
+
dz
dx
dy
dz
x2
y2
z2
= 2
= 2
=
x2 (y − z)
y (z − x)
z (x − y)
0
1
1
1
dx + 2 dy + 2 dz = 0
2
x
y
z
Integrating,
∴
(P P Savani University)
Partial Differential Equation
March 16, 2022
23 / 55
Linear PDEs
Taking 1/x2 , 1/y 2 and 1/z 2 as multipliers,
1
1
1
dx
+
dy
+
dz
dx
dy
dz
x2
y2
z2
= 2
= 2
=
x2 (y − z)
y (z − x)
z (x − y)
0
1
1
1
dx + 2 dy + 2 dz = 0
2
x
y
z
Integrating,
∴
1 1 1
+ + =b
x y z
(P P Savani University)
Partial Differential Equation
March 16, 2022
23 / 55
Linear PDEs
Taking 1/x2 , 1/y 2 and 1/z 2 as multipliers,
1
1
1
dx
+
dy
+
dz
dx
dy
dz
x2
y2
z2
= 2
= 2
=
x2 (y − z)
y (z − x)
z (x − y)
0
1
1
1
dx + 2 dy + 2 dz = 0
2
x
y
z
Integrating,
∴
1 1 1
+ + =b
x y z
Hence, the general solution is
(P P Savani University)
Partial Differential Equation
March 16, 2022
23 / 55
Linear PDEs
Taking 1/x2 , 1/y 2 and 1/z 2 as multipliers,
1
1
1
dx
+
dy
+
dz
dx
dy
dz
x2
y2
z2
= 2
= 2
=
x2 (y − z)
y (z − x)
z (x − y)
0
1
1
1
dx + 2 dy + 2 dz = 0
2
x
y
z
Integrating,
∴
1 1 1
+ + =b
x y z
Hence, the general solution is
xyz = f
1 1 1
+ +
x y z
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
23 / 55
Linear PDEs
Examples for Practice
Solve the following PDEs:
1
2
3
4
5
pz − qz = z 2 + (x + y)2
x(y − z)p + y(z − x)q = z(x − y)
y 2 zp
+ zxq = y 2
x
xzp + yzq = xy
(y 2 + z 2 )p − xyq + zx = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
24 / 55
Nonlinear PDEs
Nonlinear Partial Differential Equations of First Order
A PDE of first order is said to be nonlinear if p and q have degree more than one.
The complete solution of a nonlinear equation is given by
f (x, y, z, a, b) = 0
where a and b are two arbitrary constants. Four standard forms of these equations
are as follows:
1
f (p, q) = 0
2
f (z, p, q) = 0
3
f (x, p) = g(y, q)
4
Clairaut Equation z = px + qy + f (p, q)
(P P Savani University)
Partial Differential Equation
March 16, 2022
25 / 55
Nonlinear PDEs
Form I
Form I: f (p, q) = 0
Equations containing p and q only. Let the equation be in the form of f (p, q) = 0.
The complete solution is
z = ax + by + c
where b = φ(a), i.e., a and b satisfy the equation f (a, b) = 0.
(P P Savani University)
Partial Differential Equation
March 16, 2022
26 / 55
Nonlinear PDEs
Form I
Form I: f (p, q) = 0
Equations containing p and q only. Let the equation be in the form of f (p, q) = 0.
The complete solution is
z = ax + by + c
where b = φ(a), i.e., a and b satisfy the equation f (a, b) = 0.
√
√
Example: Solve p + q = 1
(P P Savani University)
Partial Differential Equation
March 16, 2022
26 / 55
Nonlinear PDEs
Form I
Form I: f (p, q) = 0
Equations containing p and q only. Let the equation be in the form of f (p, q) = 0.
The complete solution is
z = ax + by + c
where b = φ(a), i.e., a and b satisfy the equation f (a, b) = 0.
√
√
Example: Solve p + q = 1
Solution: The equation is of the form f (p, q) = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
26 / 55
Nonlinear PDEs
Form I
Form I: f (p, q) = 0
Equations containing p and q only. Let the equation be in the form of f (p, q) = 0.
The complete solution is
z = ax + by + c
where b = φ(a), i.e., a and b satisfy the equation f (a, b) = 0.
√
√
Example: Solve p + q = 1
Solution: The equation is of the form f (p, q) = 0
√
√
f (p, q) = p + q − 1 = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
26 / 55
Nonlinear PDEs
Form I
Form I: f (p, q) = 0
Equations containing p and q only. Let the equation be in the form of f (p, q) = 0.
The complete solution is
z = ax + by + c
where b = φ(a), i.e., a and b satisfy the equation f (a, b) = 0.
√
√
Example: Solve p + q = 1
Solution: The equation is of the form f (p, q) = 0
√
√
f (p, q) = p + q − 1 = 0
The complete solution is z = ax + by + c
(P P Savani University)
Partial Differential Equation
March 16, 2022
26 / 55
Nonlinear PDEs
Form I
Form I: f (p, q) = 0
Equations containing p and q only. Let the equation be in the form of f (p, q) = 0.
The complete solution is
z = ax + by + c
where b = φ(a), i.e., a and b satisfy the equation f (a, b) = 0.
√
√
Example: Solve p + q = 1
Solution: The equation is of the form f (p, q) = 0
√
√
f (p, q) = p + q − 1 = 0
The complete solution is z = ax + by + c
where a and b satisfy the equation f (a, b) = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
26 / 55
Nonlinear PDEs
Form I
Form I: f (p, q) = 0
Equations containing p and q only. Let the equation be in the form of f (p, q) = 0.
The complete solution is
z = ax + by + c
where b = φ(a), i.e., a and b satisfy the equation f (a, b) = 0.
√
√
Example: Solve p + q = 1
Solution: The equation is of the form f (p, q) = 0
√
√
√
√
f (p, q) = p + q − 1 = 0
a+ b−1=0
The complete solution is z = ax + by + c
where a and b satisfy the equation f (a, b) = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
26 / 55
Nonlinear PDEs
Form I
Form I: f (p, q) = 0
Equations containing p and q only. Let the equation be in the form of f (p, q) = 0.
The complete solution is
z = ax + by + c
where b = φ(a), i.e., a and b satisfy the equation f (a, b) = 0.
√
√
Example: Solve p + q = 1
Solution: The equation is of the form f (p, q) = 0
√
√
√
√
f (p, q) = p + q − 1 = 0
a+ b−1=0
√
√
The complete solution is z = ax + by + c
b=1− a
where a and b satisfy the equation f (a, b) = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
26 / 55
Nonlinear PDEs
Form I
Form I: f (p, q) = 0
Equations containing p and q only. Let the equation be in the form of f (p, q) = 0.
The complete solution is
z = ax + by + c
where b = φ(a), i.e., a and b satisfy the equation f (a, b) = 0.
√
√
Example: Solve p + q = 1
Solution: The equation is of the form f (p, q) = 0
√
√
√
√
f (p, q) = p + q − 1 = 0
a+ b−1=0
√
√
The complete solution is z = ax + by + c
b=1− a
√
where a and b satisfy the equation f (a, b) = 0
b = (1 − a)2
(P P Savani University)
Partial Differential Equation
March 16, 2022
26 / 55
Nonlinear PDEs
Form I
Form I: f (p, q) = 0
Equations containing p and q only. Let the equation be in the form of f (p, q) = 0.
The complete solution is
z = ax + by + c
where b = φ(a), i.e., a and b satisfy the equation f (a, b) = 0.
√
√
Example: Solve p + q = 1
Solution: The equation is of the form f (p, q) = 0
√
√
√
√
f (p, q) = p + q − 1 = 0
a+ b−1=0
√
√
The complete solution is z = ax + by + c
b=1− a
√
where a and b satisfy the equation f (a, b) = 0
b = (1 − a)2
√
Hence, the complete solution is z = ax + (1 − a)2 y + c
(P P Savani University)
Partial Differential Equation
March 16, 2022
J
26 / 55
Nonlinear PDEs
Form I
Example: Solve x2 p2 + y 2 q 2 = z 2
(P P Savani University)
Partial Differential Equation
March 16, 2022
27 / 55
Nonlinear PDEs
Form I
Example: Solve x2 p2 + y 2 q 2 = z 2
Solution: Rewriting the equation,
(P P Savani University)
Partial Differential Equation
March 16, 2022
27 / 55
Nonlinear PDEs
Form I
Example: Solve x2 p2 + y 2 q 2 = z 2
Solution: Rewriting the equation,
2 2
x ∂z
y ∂z
.
+
.
=1
z ∂x
z ∂y
(P P Savani University)
Partial Differential Equation
March 16, 2022
27 / 55
Nonlinear PDEs
Form I
Example: Solve x2 p2 + y 2 q 2 = z 2
Solution: Rewriting the equation,
2 2
x ∂z
y ∂z
.
+
.
=1
z ∂x
z ∂y
Let
∂x
= ∂u,
x
∂y
= ∂v,
y
(P P Savani University)
∂z
= ∂w
z
Partial Differential Equation
March 16, 2022
27 / 55
Nonlinear PDEs
Form I
Example: Solve x2 p2 + y 2 q 2 = z 2
Solution: Rewriting the equation,
2 2
x ∂z
y ∂z
.
+
.
=1
z ∂x
z ∂y
Let
∂x
∂y
∂z
= ∂u,
= ∂v,
= ∂w
x
y
z
log x = u, log y = v, log z = w
(P P Savani University)
Partial Differential Equation
March 16, 2022
27 / 55
Nonlinear PDEs
Form I
Example: Solve x2 p2 + y 2 q 2 = z 2
Solution: Rewriting the equation,
2 2
x ∂z
y ∂z
.
+
.
=1
z ∂x
z ∂y
Let
∂x
∂y
∂z
= ∂u,
= ∂v,
= ∂w
x
y
z
log x = u, log y = v, log z = w
Differentiating log z = w w.r.t. x,
(P P Savani University)
Partial Differential Equation
March 16, 2022
27 / 55
Nonlinear PDEs
Form I
Example: Solve x2 p2 + y 2 q 2 = z 2
Solution: Rewriting the equation,
2 2
x ∂z
y ∂z
.
+
.
=1
z ∂x
z ∂y
Let
∂x
∂y
∂z
= ∂u,
= ∂v,
= ∂w
x
y
z
log x = u, log y = v, log z = w
Differentiating log z = w w.r.t. x,
1 ∂z
∂w
∂w ∂u
∂w 1
=
=
·
=
·
z ∂x
∂x
∂u ∂x
∂u x
(P P Savani University)
Partial Differential Equation
March 16, 2022
27 / 55
Nonlinear PDEs
Form I
Example: Solve x2 p2 + y 2 q 2 = z 2
Solution: Rewriting the equation,
2 2
x ∂z
y ∂z
.
+
.
=1
z ∂x
z ∂y
Let
∂x
∂y
∂z
= ∂u,
= ∂v,
= ∂w
x
y
z
log x = u, log y = v, log z = w
Differentiating log z = w w.r.t. x,
1 ∂z
∂w
∂w ∂u
∂w 1
=
=
·
=
·
z ∂x
∂x
∂u ∂x
∂u x
x ∂z
∂w
=
z ∂x
∂u
(P P Savani University)
Partial Differential Equation
March 16, 2022
27 / 55
Nonlinear PDEs
Form I
Example: Solve x2 p2 + y 2 q 2 = z 2
Solution: Rewriting the equation,
2 2
x ∂z
y ∂z
.
+
.
=1
z ∂x
z ∂y
Let
∂x
∂y
∂z
= ∂u,
= ∂v,
= ∂w
x
y
z
log x = u, log y = v, log z = w
Differentiating log z = w w.r.t. x,
1 ∂z
∂w
∂w ∂u
∂w 1
=
=
·
=
·
z ∂x
∂x
∂u ∂x
∂u x
x ∂z
∂w
=
z ∂x
∂u
(P P Savani University)
Similarly,
Differentiating log z = w w.r.t. y,
Partial Differential Equation
March 16, 2022
27 / 55
Nonlinear PDEs
Form I
Example: Solve x2 p2 + y 2 q 2 = z 2
Solution: Rewriting the equation,
2 2
x ∂z
y ∂z
.
+
.
=1
z ∂x
z ∂y
Let
∂x
∂y
∂z
= ∂u,
= ∂v,
= ∂w
x
y
z
log x = u, log y = v, log z = w
Differentiating log z = w w.r.t. x,
1 ∂z
∂w
∂w ∂u
∂w 1
=
=
·
=
·
z ∂x
∂x
∂u ∂x
∂u x
x ∂z
∂w
=
z ∂x
∂u
(P P Savani University)
Similarly,
Differentiating log z = w w.r.t. y,
Partial Differential Equation
y ∂z
∂w
=
z ∂y
∂v
March 16, 2022
27 / 55
Nonlinear PDEs
Form I
Substituting the values in the equation,
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
28 / 55
Nonlinear PDEs
Form I
Substituting the values in the equation,
∂w
∂u
2
+
∂w
∂v
2
=1
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
28 / 55
Nonlinear PDEs
Form I
Substituting the values in the equation,
∂w
∂u
2
+
∂w
∂v
2
=1
P 2 + Q2 = 1
where P =
∂w
∂u
and Q =
∂w
∂v
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
28 / 55
Nonlinear PDEs
Form I
Substituting the values in the equation,
∂w
∂u
2
+
∂w
∂v
2
=1
P 2 + Q2 = 1
where P =
∂w
∂u
and Q =
∂w
∂v
The equation is of the form f (P, Q) = 0.
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
28 / 55
Nonlinear PDEs
Form I
Substituting the values in the equation,
∂w
∂u
2
+
∂w
∂v
2
=1
P 2 + Q2 = 1
where P =
∂w
∂u
and Q =
∂w
∂v
The equation is of the form f (P, Q) = 0.
f (P, Q) = P 2 + Q2 − 1
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
28 / 55
Nonlinear PDEs
Form I
Substituting the values in the equation,
∂w
∂u
2
+
∂w
∂v
2
=1
P 2 + Q2 = 1
where P =
∂w
∂u
and Q =
∂w
∂v
The equation is of the form f (P, Q) = 0.
f (P, Q) = P 2 + Q2 − 1
The complete solution is w = au + bv + c
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
28 / 55
Nonlinear PDEs
Form I
Substituting the values in the equation,
∂w
∂u
2
+
∂w
∂v
2
=1
P 2 + Q2 = 1
where P =
∂w
∂u
and Q =
∂w
∂v
The equation is of the form f (P, Q) = 0.
f (P, Q) = P 2 + Q2 − 1
The complete solution is w = au + bv + c
√
where a2 + b2 − 1 = 0
or
b = 1 − a2
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
28 / 55
Nonlinear PDEs
Form I
Substituting the values in the equation,
∂w
∂u
2
+
∂w
∂v
2
=1
P 2 + Q2 = 1
where P =
∂w
∂u
and Q =
∂w
∂v
The equation is of the form f (P, Q) = 0.
f (P, Q) = P 2 + Q2 − 1
The complete solution is w = au + bv + c
√
where a2 + b2 − 1 = 0
or
b = 1 − a2
p
∴ w = au + ( 1 − a2 )v + c
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
28 / 55
Nonlinear PDEs
Form I
Substituting the values in the equation,
∂w
∂u
2
+
∂w
∂v
2
=1
P 2 + Q2 = 1
where P =
∂w
∂u
and Q =
∂w
∂v
The equation is of the form f (P, Q) = 0.
f (P, Q) = P 2 + Q2 − 1
The complete solution is w = au + bv + c
√
where a2 + b2 − 1 = 0
or
b = 1 − a2
p
∴ w = au + ( 1 − a2 )v + c
p
log z = a log x + 1 − a2 log y + c
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
28 / 55
Nonlinear PDEs
Form II
Form II: f (z, p, q) = 0
Equations not containing x and y. Let the equation be in the form of f (z, p, q) = 0
Assuming q = ap, the above equation reduces to f (z, p, ap) = 0
Solving for p, we derive p = φ(z)
Now,
dz = pdx + qdy
= pdx + apdy
= p(dx + ady)
= φ(z)(dx + ady)
Z
dz
= x + ay + b
φ(z)
which gives the complete solution, where a and b are arbitrary constants.
Integrating,
(P P Savani University)
Partial Differential Equation
March 16, 2022
29 / 55
Nonlinear PDEs
Form II
Example: Solve z 2 (p2 z 2 + q 2 ) = 1
(P P Savani University)
Partial Differential Equation
March 16, 2022
30 / 55
Nonlinear PDEs
Form II
Example: Solve z 2 (p2 z 2 + q 2 ) = 1
Solution: Putting q = ap,
(P P Savani University)
Partial Differential Equation
March 16, 2022
30 / 55
Nonlinear PDEs
Form II
Example: Solve z 2 (p2 z 2 + q 2 ) = 1
Solution: Putting q = ap,
z 2 (p2 z 2 + a2 p2 ) = 1
(P P Savani University)
Partial Differential Equation
March 16, 2022
30 / 55
Nonlinear PDEs
Form II
Example: Solve z 2 (p2 z 2 + q 2 ) = 1
Solution: Putting q = ap,
z 2 (p2 z 2 + a2 p2 ) = 1
p2 =
(P P Savani University)
1
+ a2 )
z 2 (z 2
Partial Differential Equation
March 16, 2022
30 / 55
Nonlinear PDEs
Form II
Example: Solve z 2 (p2 z 2 + q 2 ) = 1
Solution: Putting q = ap,
z 2 (p2 z 2 + a2 p2 ) = 1
1
+ a2 )
1
p= √
= φ(z)
z z 2 + a2
p2 =
(P P Savani University)
z 2 (z 2
Partial Differential Equation
March 16, 2022
30 / 55
Nonlinear PDEs
Form II
Example: Solve z 2 (p2 z 2 + q 2 ) = 1
Solution: Putting q = ap,
z 2 (p2 z 2 + a2 p2 ) = 1
1
+ a2 )
1
p= √
= φ(z)
z z 2 + a2
p2 =
z 2 (z 2
Now,
(P P Savani University)
Partial Differential Equation
March 16, 2022
30 / 55
Nonlinear PDEs
Form II
Example: Solve z 2 (p2 z 2 + q 2 ) = 1
Solution: Putting q = ap,
z 2 (p2 z 2 + a2 p2 ) = 1
1
+ a2 )
1
p= √
= φ(z)
z z 2 + a2
p2 =
Now,
z 2 (z 2
dz = pdx + qdy = pdx + apdy
(P P Savani University)
Partial Differential Equation
March 16, 2022
30 / 55
Nonlinear PDEs
Form II
Example: Solve z 2 (p2 z 2 + q 2 ) = 1
Solution: Putting q = ap,
z 2 (p2 z 2 + a2 p2 ) = 1
1
+ a2 )
1
p= √
= φ(z)
z z 2 + a2
p2 =
Now,
z 2 (z 2
dz = pdx + qdy = pdx + apdy
= p(dx + ady)
(P P Savani University)
Partial Differential Equation
March 16, 2022
30 / 55
Nonlinear PDEs
Form II
Example: Solve z 2 (p2 z 2 + q 2 ) = 1
Solution: Putting q = ap,
z 2 (p2 z 2 + a2 p2 ) = 1
1
+ a2 )
1
p= √
= φ(z)
z z 2 + a2
p2 =
Now,
z 2 (z 2
dz = pdx + qdy = pdx + apdy
= p(dx + ady)
1
= p
(dx + ady)
z (z 2 + a2 )
(P P Savani University)
Partial Differential Equation
March 16, 2022
30 / 55
Nonlinear PDEs
Form II
Example: Solve z 2 (p2 z 2 + q 2 ) = 1
Solution: Putting q = ap,
z 2 (p2 z 2 + a2 p2 ) = 1
1
+ a2 )
1
p= √
= φ(z)
z z 2 + a2
p2 =
Now,
z 2 (z 2
dz = pdx + qdy = pdx + apdy
= p(dx + ady)
1
= p
(dx + ady)
z (z 2 + a2 )
√
z z 2 + a2 dz = dx + ady
(P P Savani University)
Partial Differential Equation
March 16, 2022
30 / 55
Nonlinear PDEs
Form II
Integrating,
(P P Savani University)
Partial Differential Equation
March 16, 2022
31 / 55
Nonlinear PDEs
Form II
Integrating,
Z
1 2z
(z 2 + a2 ) /2 dz = x + ay + b
2
(P P Savani University)
Partial Differential Equation
March 16, 2022
31 / 55
Nonlinear PDEs
Integrating,
Z
1 2z
(z 2 + a2 ) /2 dz = x + ay + b
2
2
1 (z + a2 )3/2
= x + ay + b
3
2
2
(P P Savani University)
Form II
Z
[f (x)]n+1
n 0
∵ [f (x)] f (x)dx =
n+1
Partial Differential Equation
March 16, 2022
31 / 55
Nonlinear PDEs
Integrating,
Z
1 2z
(z 2 + a2 ) /2 dz = x + ay + b
2
2
1 (z + a2 )3/2
= x + ay + b
3
2
2
Form II
Z
[f (x)]n+1
n 0
∵ [f (x)] f (x)dx =
n+1
3
(z 2 + a2 ) /2 = 3(x + ay + b)
(P P Savani University)
Partial Differential Equation
March 16, 2022
31 / 55
Nonlinear PDEs
Integrating,
Z
1 2z
(z 2 + a2 ) /2 dz = x + ay + b
2
2
1 (z + a2 )3/2
= x + ay + b
3
2
2
Form II
Z
[f (x)]n+1
n 0
∵ [f (x)] f (x)dx =
n+1
3
(z 2 + a2 ) /2 = 3(x + ay + b)
(z 2 + a2 )3 = 9(x + ay + b)2
(P P Savani University)
Partial Differential Equation
March 16, 2022
31 / 55
Nonlinear PDEs
Integrating,
Z
1 2z
(z 2 + a2 ) /2 dz = x + ay + b
2
2
1 (z + a2 )3/2
= x + ay + b
3
2
2
Form II
Z
[f (x)]n+1
n 0
∵ [f (x)] f (x)dx =
n+1
3
(z 2 + a2 ) /2 = 3(x + ay + b)
(z 2 + a2 )3 = 9(x + ay + b)2
which gives the complete solution of the given equation.
(P P Savani University)
Partial Differential Equation
J
March 16, 2022
31 / 55
Nonlinear PDEs
Form III
Form III: f (x, p) = F (y, q)
Equations in which z is absent and the terms containing x and p can be separated
from those containing y and q. Let the equation be in the form of f (x, p) = F (y, q)
As a trial solution assume that f (x, p) = F (y, q) = a, say
Let p = φ(x) and q = ψ(y).
Now,
dz = pdx + qdy = φ(x)dx + ψ(y)dy
Integrating,
Z
z=
Z
φ(x)dx +
ψ(y)dy + b
which is the desired complete solution containing two arbitrary constants a and b.
(P P Savani University)
Partial Differential Equation
March 16, 2022
32 / 55
Nonlinear PDEs
Form III
Example: Solve yp = 2yx + log q
(P P Savani University)
Partial Differential Equation
March 16, 2022
33 / 55
Nonlinear PDEs
Form III
Example: Solve yp = 2yx + log q
Solution: Dividing the equation by y,
(P P Savani University)
Partial Differential Equation
March 16, 2022
33 / 55
Nonlinear PDEs
Form III
Example: Solve yp = 2yx + log q
Solution: Dividing the equation by y,
p = 2x +
(P P Savani University)
1
log q
y
Partial Differential Equation
March 16, 2022
33 / 55
Nonlinear PDEs
Form III
Example: Solve yp = 2yx + log q
Solution: Dividing the equation by y,
p = 2x +
p − 2x =
(P P Savani University)
1
log q
y
1
log q
y
Partial Differential Equation
March 16, 2022
33 / 55
Nonlinear PDEs
Form III
Example: Solve yp = 2yx + log q
Solution: Dividing the equation by y,
p = 2x +
p − 2x =
Let
1
log q
y
1
log q
y
1
log q = a
y
p − 2x = a,
(P P Savani University)
Partial Differential Equation
March 16, 2022
33 / 55
Nonlinear PDEs
Form III
Example: Solve yp = 2yx + log q
Solution: Dividing the equation by y,
p = 2x +
p − 2x =
Let
1
log q
y
1
log q
y
1
log q = a
y
p − 2x = a,
p = 2x + a
log q = ay
q = eay
(P P Savani University)
Partial Differential Equation
March 16, 2022
33 / 55
Nonlinear PDEs
Form III
Example: Solve yp = 2yx + log q
Solution: Dividing the equation by y,
p = 2x +
p − 2x =
Let
1
log q
y
1
log q
y
1
log q = a
y
p − 2x = a,
p = 2x + a
log q = ay
q = eay
Now,
dz = pdx + qdy
(P P Savani University)
Partial Differential Equation
March 16, 2022
33 / 55
Nonlinear PDEs
Form III
Example: Solve yp = 2yx + log q
Solution: Dividing the equation by y,
p = 2x +
p − 2x =
Let
1
log q
y
1
log q
y
1
log q = a
y
p − 2x = a,
p = 2x + a
log q = ay
q = eay
Now,
dz = pdx + qdy
= (2x + a)dx + eay dy
(P P Savani University)
Partial Differential Equation
March 16, 2022
33 / 55
Nonlinear PDEs
Form III
Example: Solve yp = 2yx + log q
Solution: Dividing the equation by y,
p = 2x +
p − 2x =
Let
1
log q
y
1
log q
y
1
log q = a
y
p − 2x = a,
p = 2x + a
log q = ay
q = eay
Now,
dz = pdx + qdy
= (2x + a)dx + eay dy
Integrating,
(P P Savani University)
Partial Differential Equation
March 16, 2022
33 / 55
Nonlinear PDEs
Form III
Example: Solve yp = 2yx + log q
Solution: Dividing the equation by y,
p = 2x +
p − 2x =
Let
1
log q
y
1
log q
y
1
log q = a
y
p − 2x = a,
p = 2x + a
log q = ay
q = eay
Now,
dz = pdx + qdy
= (2x + a)dx + eay dy
Integrating,
(P P Savani University)
z = x2 + ax +
eay
+b
a
Partial Differential Equation
March 16, 2022
33 / 55
Nonlinear PDEs
Form III
Example: Solve yp = 2yx + log q
Solution: Dividing the equation by y,
p = 2x +
p − 2x =
Let
1
log q
y
1
log q
y
1
log q = a
y
p − 2x = a,
p = 2x + a
log q = ay
q = eay
Now,
dz = pdx + qdy
= (2x + a)dx + eay dy
Integrating,
eay
+b
a
az = ax2 + a2 x + eay + ab
z = x2 + ax +
which gives the complete solution of the given equation.
(P P Savani University)
Partial Differential Equation
J
March 16, 2022
33 / 55
Nonlinear PDEs
Form III
Example: Solve zpy 2 = x(y 2 + z 2 q 2 )
Solution:
(2)
(P P Savani University)
Partial Differential Equation
March 16, 2022
34 / 55
Nonlinear PDEs
Form III
Example: Solve zpy 2 = x(y 2 + z 2 q 2 )
Solution:
zpy 2 = x(y 2 + z 2 q 2 )
(1)
(2)
(P P Savani University)
Partial Differential Equation
March 16, 2022
34 / 55
Nonlinear PDEs
Form III
Example: Solve zpy 2 = x(y 2 + z 2 q 2 )
Solution:
zpy 2 = x(y 2 + z 2 q 2 )
Let,
(1)
z∂z = ∂Z
(2)
(P P Savani University)
Partial Differential Equation
March 16, 2022
34 / 55
Nonlinear PDEs
Form III
Example: Solve zpy 2 = x(y 2 + z 2 q 2 )
Solution:
zpy 2 = x(y 2 + z 2 q 2 )
Let,
(1)
z∂z = ∂Z
z2
=Z
2
(P P Savani University)
Partial Differential Equation
(2)
March 16, 2022
34 / 55
Nonlinear PDEs
Form III
Example: Solve zpy 2 = x(y 2 + z 2 q 2 )
Solution:
zpy 2 = x(y 2 + z 2 q 2 )
Let,
(1)
z∂z = ∂Z
z2
=Z
2
(2)
Partially Differentiating Eq. (2) w.r.t. x,
(P P Savani University)
Partial Differential Equation
March 16, 2022
34 / 55
Nonlinear PDEs
Form III
Example: Solve zpy 2 = x(y 2 + z 2 q 2 )
Solution:
zpy 2 = x(y 2 + z 2 q 2 )
Let,
(1)
z∂z = ∂Z
z2
=Z
2
(2)
Partially Differentiating Eq. (2) w.r.t. x,
z
(P P Savani University)
∂z
∂Z
=
= P, say
∂x
∂x
Partial Differential Equation
March 16, 2022
34 / 55
Nonlinear PDEs
Form III
Example: Solve zpy 2 = x(y 2 + z 2 q 2 )
Solution:
zpy 2 = x(y 2 + z 2 q 2 )
Let,
(1)
z∂z = ∂Z
z2
=Z
2
(2)
Partially Differentiating Eq. (2) w.r.t. x,
z
(P P Savani University)
∂z
∂Z
=
= P, say
∂x
∂x
zp = P
Partial Differential Equation
March 16, 2022
34 / 55
Nonlinear PDEs
Form III
Example: Solve zpy 2 = x(y 2 + z 2 q 2 )
Solution:
zpy 2 = x(y 2 + z 2 q 2 )
Let,
(1)
z∂z = ∂Z
z2
=Z
2
(2)
Partially Differentiating Eq. (2) w.r.t. x,
z
∂z
∂Z
=
= P, say
∂x
∂x
zp = P
Partially Differentiating Eq. (2) w.r.t. y,
(P P Savani University)
Partial Differential Equation
March 16, 2022
34 / 55
Nonlinear PDEs
Form III
Example: Solve zpy 2 = x(y 2 + z 2 q 2 )
Solution:
zpy 2 = x(y 2 + z 2 q 2 )
Let,
(1)
z∂z = ∂Z
z2
=Z
2
(2)
Partially Differentiating Eq. (2) w.r.t. x,
z
∂z
∂Z
=
= P, say
∂x
∂x
zp = P
Partially Differentiating Eq. (2) w.r.t. y,
z
(P P Savani University)
∂z
∂Z
=
=Q
∂y
∂y
Partial Differential Equation
March 16, 2022
34 / 55
Nonlinear PDEs
Form III
Example: Solve zpy 2 = x(y 2 + z 2 q 2 )
Solution:
zpy 2 = x(y 2 + z 2 q 2 )
Let,
(1)
z∂z = ∂Z
z2
=Z
2
(2)
Partially Differentiating Eq. (2) w.r.t. x,
z
∂z
∂Z
=
= P, say
∂x
∂x
zp = P
Partially Differentiating Eq. (2) w.r.t. y,
z
∂z
∂Z
=
=Q
∂y
∂y
zq = Q
(P P Savani University)
Partial Differential Equation
March 16, 2022
34 / 55
Nonlinear PDEs
Form III
Substituting in equation (1),
(P P Savani University)
Partial Differential Equation
March 16, 2022
35 / 55
Nonlinear PDEs
Form III
Substituting in equation (1),
P y 2 = x(y 2 + Q2 )
(P P Savani University)
Partial Differential Equation
March 16, 2022
35 / 55
Nonlinear PDEs
Form III
Substituting in equation (1),
P y 2 = x(y 2 + Q2 )
P
Q2 + y 2
=
x
y2
(P P Savani University)
Partial Differential Equation
March 16, 2022
35 / 55
Nonlinear PDEs
Form III
Substituting in equation (1),
P y 2 = x(y 2 + Q2 )
P
Q2 + y 2
=
x
y2
The equation is in the form f (x, P ) = g(y, Q).
(P P Savani University)
Partial Differential Equation
March 16, 2022
35 / 55
Nonlinear PDEs
Form III
Substituting in equation (1),
P y 2 = x(y 2 + Q2 )
P
Q2 + y 2
=
x
y2
The equation is in the form f (x, P ) = g(y, Q).
Let
P
= a,
x
(P P Savani University)
Q2 + y 2
=a
y2
Partial Differential Equation
March 16, 2022
35 / 55
Nonlinear PDEs
Form III
Substituting in equation (1),
P y 2 = x(y 2 + Q2 )
P
Q2 + y 2
=
x
y2
The equation is in the form f (x, P ) = g(y, Q).
Let
P
= a,
x
P = ax
(P P Savani University)
Q2 + y 2
=a
y2
√
Q=y a−1
Partial Differential Equation
March 16, 2022
35 / 55
Nonlinear PDEs
Form III
Substituting in equation (1),
P y 2 = x(y 2 + Q2 )
P
Q2 + y 2
=
x
y2
The equation is in the form f (x, P ) = g(y, Q).
Let
Q2 + y 2
=a
y2
√
Q=y a−1
P
= a,
x
P = ax
Now,
dZ =
(P P Savani University)
∂Z
∂Z
dx +
dy
∂x
∂y
Partial Differential Equation
March 16, 2022
35 / 55
Nonlinear PDEs
Form III
Substituting in equation (1),
P y 2 = x(y 2 + Q2 )
P
Q2 + y 2
=
x
y2
The equation is in the form f (x, P ) = g(y, Q).
Let
Q2 + y 2
=a
y2
√
Q=y a−1
P
= a,
x
P = ax
Now,
dZ =
∂Z
∂Z
dx +
dy
∂x
∂y
zdz = P dx + Qdy
(P P Savani University)
Partial Differential Equation
March 16, 2022
35 / 55
Nonlinear PDEs
Form III
Substituting in equation (1),
P y 2 = x(y 2 + Q2 )
P
Q2 + y 2
=
x
y2
The equation is in the form f (x, P ) = g(y, Q).
Let
Q2 + y 2
=a
y2
√
Q=y a−1
P
= a,
x
P = ax
Now,
dZ =
∂Z
∂Z
dx +
dy
∂x
∂y
zdz = P dx + Qdy
√
= axdx + y a − 1dy
(P P Savani University)
Partial Differential Equation
March 16, 2022
35 / 55
Nonlinear PDEs
Form III
Substituting in equation (1),
P y 2 = x(y 2 + Q2 )
P
Q2 + y 2
=
x
y2
The equation is in the form f (x, P ) = g(y, Q).
Let
Q2 + y 2
=a
y2
√
Q=y a−1
P
= a,
x
P = ax
Now,
dZ =
∂Z
∂Z
dx +
dy
∂x
∂y
zdz = P dx + Qdy
√
= axdx + y a − 1dy
Integrating,
(P P Savani University)
z2
x2 y 2 √
b
=a +
a−1+
2
2
2
2
Partial Differential Equation
March 16, 2022
35 / 55
Nonlinear PDEs
Form III
Substituting in equation (1),
P y 2 = x(y 2 + Q2 )
P
Q2 + y 2
=
x
y2
The equation is in the form f (x, P ) = g(y, Q).
Let
Q2 + y 2
=a
y2
√
Q=y a−1
P
= a,
x
P = ax
Now,
dZ =
∂Z
∂Z
dx +
dy
∂x
∂y
zdz = P dx + Qdy
√
= axdx + y a − 1dy
Integrating,
z2
x2 y 2 √
b
=a +
a−1+
2
2
2
2
√
z 2 = ax2 + y 2 a − 1 + b
which gives the complete solution of the given equation.
(P P Savani University)
Partial Differential Equation
J
March 16, 2022
35 / 55
Nonlinear PDEs
Form IV
Form IV: (Clairaut Equation)
Let the equation be z = px + qy + f (p, q)
The complete solution of this equation is
z = ax + by + f (a, b)
which is obtained by replacing p by a and q by b.
(P P Savani University)
Partial Differential Equation
March 16, 2022
36 / 55
Nonlinear PDEs
Form IV
Form IV: (Clairaut Equation)
Let the equation be z = px + qy + f (p, q)
The complete solution of this equation is
z = ax + by + f (a, b)
which is obtained by replacing p by a and q by b.
√
Example: Solve z = px + qy − 2 pq
(P P Savani University)
Partial Differential Equation
March 16, 2022
36 / 55
Nonlinear PDEs
Form IV
Form IV: (Clairaut Equation)
Let the equation be z = px + qy + f (p, q)
The complete solution of this equation is
z = ax + by + f (a, b)
which is obtained by replacing p by a and q by b.
√
Example: Solve z = px + qy − 2 pq
Solution: The given equation is of the form
z = px + qy + f (p, q)
(P P Savani University)
Partial Differential Equation
March 16, 2022
36 / 55
Nonlinear PDEs
Form IV
Form IV: (Clairaut Equation)
Let the equation be z = px + qy + f (p, q)
The complete solution of this equation is
z = ax + by + f (a, b)
which is obtained by replacing p by a and q by b.
√
Example: Solve z = px + qy − 2 pq
Solution: The given equation is of the form
z = px + qy + f (p, q)
Hence, the complete solution is obtained by replacing p by a and q by b in the given
equation.
(P P Savani University)
Partial Differential Equation
March 16, 2022
36 / 55
Nonlinear PDEs
Form IV
Form IV: (Clairaut Equation)
Let the equation be z = px + qy + f (p, q)
The complete solution of this equation is
z = ax + by + f (a, b)
which is obtained by replacing p by a and q by b.
√
Example: Solve z = px + qy − 2 pq
Solution: The given equation is of the form
z = px + qy + f (p, q)
Hence, the complete solution is obtained by replacing p by a and q by b in the given
√
equation.
z = ax + by − 2 ab
(P P Savani University)
Partial Differential Equation
March 16, 2022
J
36 / 55
Nonlinear PDEs
Form IV
Examples for Practice
Solve the following PDEs:
(Non Linear Form I)
(Non Linear Form III)
1
p + q2 = 1
1
p2 + q 2 = x + y
2
p2 + q 2 = 1
2
p2 − q 2 = x − y
3
p2 + q 2 = npq
3
p2 + q 2 = z 2 (x + y)
(Non Linear Form II)
(Non Linear Form IV)
1
p3 + q 3 = 27z
1
2
p(1 + q) = qz
2
3
q 2 y 2 = z(z − px)
3
(P P Savani University)
z = px + qy + p2 q 2
p
z = px + qy + c 1 + p2 + q 2
(p − q)(z − px − qy) = 1
Partial Differential Equation
March 16, 2022
37 / 55
Homogeneous Linear PDEs of Higher Order
Homogeneous Linear PDEs of Higher Order with Constant
Coefficients
An equation of the form
a0
∂nz
∂nz
∂nz
+
·
·
·
+
a
+
a
= F (x, y)
1
n
∂xn
∂xn−1 ∂y
∂y n
(3)
where a0 , a1 , · · · , an are constants, is known as a homogeneous linear partial differential equations of
nth order with constant coefficients. It is known as homogeneous because all terms in the equation contain
∂
∂
= D and ∂y
= D0 in Eq. (3),
derivatives of the same order. Replacing ∂x
(a0 Dn + a1 Dn−1 D0 + · · · + an D0n )z = F (x, y)
f (D, D0 )z = F (x, y)
As in the case of ordinary linear differential equation with constant coefficients, the complete solution of Eq. (3)
is obtained in two parts, one as a Complementary Function (CF) and the other as a Partial Integral (PI).
The complementary function is the solution of the equation f (D, D0 )z = 0.
(P P Savani University)
Partial Differential Equation
March 16, 2022
38 / 55
Homogeneous Linear PDEs of Higher Order
Complementary function
Complementary Function
Consider the equation
∂ 2z
∂ 2z
∂ 2z
+
a
+
a
=0
1
2
∂x2
∂x∂y
∂y 2
(4)
(a0 D2 + a1 DD0 + a2 D02 )z = 0
(5)
a0
which in symbolic form is
Its symbolic operator equated to zero, i.e.,
a0 m2 + a1 m + a2 = 0
(6)
is called the auxiliary equation (AE). Let m1 , m2 be the roots of Eq. (6).
Remark: The auxiliary equation is obtained by replacing D with m and D0 with 1
in the given differential equation.
If F (x, y) = 0 the particular integral = 0.
(P P Savani University)
Partial Differential Equation
March 16, 2022
39 / 55
Homogeneous Linear PDEs of Higher Order
Complementary function
Case I: Real and Distinct Roots
If m1 , m2 are real and distinct then Eq. (5) is equivalent to
(D − m1 D0 )(D − m2 D0 )z = 0
It will be satisfied by the solution of (D − m1 D0 )z = 0 i.e.,
This is a Lagrange’s linear and the subsidiary equations are
dx
dy
dz
=
=
1
−m1
0
dy + m1 dx = 0, dz = 0
Integrating, y + m1 x = a,
(7)
p − m1 q = 0
z=b
The solution of Eq. (7) is z = φ(y + m1 x)
Similarly, the solutions of the other factors of Eq. (7) are z = f (y + m2 x)
(P P Savani University)
Partial Differential Equation
March 16, 2022
40 / 55
Homogeneous Linear PDEs of Higher Order
Complementary function
Case II: Equal Roots
Let the roots be equal m1 = m2 = m then Eq. (5) is equivalent to
(D − m1 D0 )2 z = 0
i.e.,
(D − mD0 )(D − mD0 )z = 0
(8)
Putting (D − mD0 )z = u it becomes (D − mD0 )u = 0 which gives
u = φ(y + mx)
(D − mD0 )z = φ(y + mx)
p − mq = φ(y + mx)
This is again Lagrange’s linear and the subsidiary equations are
dy
dz
dx
=
=
1
−m
φ(y + mx)
(P P Savani University)
Partial Differential Equation
(9)
March 16, 2022
41 / 55
Homogeneous Linear PDEs of Higher Order
Complementary function
Taking the first and second fractions of Eq. (9),
−mdx = dy
Integrating, y + mx = a
Taking the first and third fractions of Eq. (9),
dx
dz
=
1
φ(a)
dz = φ(a)dx
Integrating, z = xφ(a) + b
= xφ(y + mx) + b
Thus, the complete solution of Eq. (8) is
z − xφ(y + mx) = f (y + mx)
(P P Savani University)
i.e.,
z = f (y + mx) + xφ(y + mx)
Partial Differential Equation
March 16, 2022
42 / 55
Homogeneous Linear PDEs of Higher Order
Example: Solve 2
∂2z
∂x2
(P P Savani University)
+5
∂2z
∂x∂y
+2
∂2z
∂y 2
Complementary function
=0
Partial Differential Equation
March 16, 2022
43 / 55
Homogeneous Linear PDEs of Higher Order
Example: Solve 2
∂2z
∂2z
Complementary function
∂2z
+5
+2 2 =0
∂x2
∂x∂y
∂y
Solution: The equation can be written as
(P P Savani University)
Partial Differential Equation
March 16, 2022
43 / 55
Homogeneous Linear PDEs of Higher Order
Example: Solve 2
∂2z
∂2z
Complementary function
∂2z
+5
+2 2 =0
∂x2
∂x∂y
∂y
Solution: The equation can be written as
(2D2 + 5DD0 + 2D02 )z = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
43 / 55
Homogeneous Linear PDEs of Higher Order
Example: Solve 2
∂2z
∂2z
Complementary function
∂2z
+5
+2 2 =0
∂x2
∂x∂y
∂y
Solution: The equation can be written as
(2D2 + 5DD0 + 2D02 )z = 0
The auxiliary equation is
(P P Savani University)
Partial Differential Equation
March 16, 2022
43 / 55
Homogeneous Linear PDEs of Higher Order
Example: Solve 2
∂2z
∂2z
Complementary function
∂2z
+5
+2 2 =0
∂x2
∂x∂y
∂y
Solution: The equation can be written as
(2D2 + 5DD0 + 2D02 )z = 0
The auxiliary equation is
2m2 + 5m + 2 = 0,
(P P Savani University)
where D = m and D0 = 1
Partial Differential Equation
March 16, 2022
43 / 55
Homogeneous Linear PDEs of Higher Order
Example: Solve 2
∂2z
∂2z
Complementary function
∂2z
+5
+2 2 =0
∂x2
∂x∂y
∂y
Solution: The equation can be written as
(2D2 + 5DD0 + 2D02 )z = 0
The auxiliary equation is
2m2 + 5m + 2 = 0,
where D = m and D0 = 1
1
m = −2, −
(real and distinct roots)
2
(P P Savani University)
Partial Differential Equation
March 16, 2022
43 / 55
Homogeneous Linear PDEs of Higher Order
Example: Solve 2
∂2z
∂2z
Complementary function
∂2z
+5
+2 2 =0
∂x2
∂x∂y
∂y
Solution: The equation can be written as
(2D2 + 5DD0 + 2D02 )z = 0
The auxiliary equation is
2m2 + 5m + 2 = 0,
where D = m and D0 = 1
1
m = −2, −
(real and distinct roots)
2
x
CF = f1 (y − 2x) + f2 (y − )
2
(P P Savani University)
Partial Differential Equation
March 16, 2022
43 / 55
Homogeneous Linear PDEs of Higher Order
Example: Solve 2
∂2z
∂2z
Complementary function
∂2z
+5
+2 2 =0
∂x2
∂x∂y
∂y
Solution: The equation can be written as
(2D2 + 5DD0 + 2D02 )z = 0
The auxiliary equation is
2m2 + 5m + 2 = 0,
where D = m and D0 = 1
1
m = −2, −
(real and distinct roots)
2
x
CF = f1 (y − 2x) + f2 (y − )
2
F (x, y) = 0
PI = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
43 / 55
Homogeneous Linear PDEs of Higher Order
Example: Solve 2
∂2z
∂2z
Complementary function
∂2z
+5
+2 2 =0
∂x2
∂x∂y
∂y
Solution: The equation can be written as
(2D2 + 5DD0 + 2D02 )z = 0
The auxiliary equation is
2m2 + 5m + 2 = 0,
where D = m and D0 = 1
1
m = −2, −
(real and distinct roots)
2
x
CF = f1 (y − 2x) + f2 (y − )
2
F (x, y) = 0
PI = 0
Hence, the complete solution is
(P P Savani University)
Partial Differential Equation
March 16, 2022
43 / 55
Homogeneous Linear PDEs of Higher Order
Example: Solve 2
∂2z
∂2z
Complementary function
∂2z
+5
+2 2 =0
∂x2
∂x∂y
∂y
Solution: The equation can be written as
(2D2 + 5DD0 + 2D02 )z = 0
The auxiliary equation is
2m2 + 5m + 2 = 0,
where D = m and D0 = 1
1
m = −2, −
(real and distinct roots)
2
x
CF = f1 (y − 2x) + f2 (y − )
2
F (x, y) = 0
PI = 0
Hence, the complete solution is
x
z = f1 (y − 2x) + f2 (y − )
2
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
43 / 55
Homogeneous Linear PDEs of Higher Order
Complementary function
Example: Solve 25r − 40s + 16t = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
44 / 55
Homogeneous Linear PDEs of Higher Order
Complementary function
Example: Solve 25r − 40s + 16t = 0
Solution: The equation can be written as
(P P Savani University)
Partial Differential Equation
March 16, 2022
44 / 55
Homogeneous Linear PDEs of Higher Order
Complementary function
Example: Solve 25r − 40s + 16t = 0
Solution: The equation can be written as
∂ 2z
∂ 2z
∂ 2z
25 2 − 40
+ 16 2 = 0
∂x
∂x∂y
∂y
(P P Savani University)
Partial Differential Equation
March 16, 2022
44 / 55
Homogeneous Linear PDEs of Higher Order
Complementary function
Example: Solve 25r − 40s + 16t = 0
Solution: The equation can be written as
∂ 2z
∂ 2z
∂ 2z
25 2 − 40
+ 16 2 = 0 ⇒ (25D2 − 40DD0 + 16D02 )z = 0
∂x
∂x∂y
∂y
(P P Savani University)
Partial Differential Equation
March 16, 2022
44 / 55
Homogeneous Linear PDEs of Higher Order
Complementary function
Example: Solve 25r − 40s + 16t = 0
Solution: The equation can be written as
∂ 2z
∂ 2z
∂ 2z
25 2 − 40
+ 16 2 = 0 ⇒ (25D2 − 40DD0 + 16D02 )z = 0
∂x
∂x∂y
∂y
The auxiliary equation is
(P P Savani University)
Partial Differential Equation
March 16, 2022
44 / 55
Homogeneous Linear PDEs of Higher Order
Complementary function
Example: Solve 25r − 40s + 16t = 0
Solution: The equation can be written as
∂ 2z
∂ 2z
∂ 2z
25 2 − 40
+ 16 2 = 0 ⇒ (25D2 − 40DD0 + 16D02 )z = 0
∂x
∂x∂y
∂y
The auxiliary equation is
25m2 − 40m + 16 = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
44 / 55
Homogeneous Linear PDEs of Higher Order
Complementary function
Example: Solve 25r − 40s + 16t = 0
Solution: The equation can be written as
∂ 2z
∂ 2z
∂ 2z
25 2 − 40
+ 16 2 = 0 ⇒ (25D2 − 40DD0 + 16D02 )z = 0
∂x
∂x∂y
∂y
The auxiliary equation is
25m2 − 40m + 16 = 0
4 4
m= ,
5 5
(P P Savani University)
(repeated roots)
Partial Differential Equation
March 16, 2022
44 / 55
Homogeneous Linear PDEs of Higher Order
Complementary function
Example: Solve 25r − 40s + 16t = 0
Solution: The equation can be written as
∂ 2z
∂ 2z
∂ 2z
25 2 − 40
+ 16 2 = 0 ⇒ (25D2 − 40DD0 + 16D02 )z = 0
∂x
∂x∂y
∂y
The auxiliary equation is
25m2 − 40m + 16 = 0
4 4
m= ,
(repeated roots)
5 5 4
4
CF = φ1 y + x + xφ2 y + x
5
5
(P P Savani University)
Partial Differential Equation
March 16, 2022
44 / 55
Homogeneous Linear PDEs of Higher Order
Complementary function
Example: Solve 25r − 40s + 16t = 0
Solution: The equation can be written as
∂ 2z
∂ 2z
∂ 2z
25 2 − 40
+ 16 2 = 0 ⇒ (25D2 − 40DD0 + 16D02 )z = 0
∂x
∂x∂y
∂y
The auxiliary equation is
25m2 − 40m + 16 = 0
4 4
m= ,
(repeated roots)
5 5 4
4
CF = φ1 y + x + xφ2 y + x = f1 (5y + 4x) + xf2 (5y + 4x)
5
5
(P P Savani University)
Partial Differential Equation
March 16, 2022
44 / 55
Homogeneous Linear PDEs of Higher Order
Complementary function
Example: Solve 25r − 40s + 16t = 0
Solution: The equation can be written as
∂ 2z
∂ 2z
∂ 2z
25 2 − 40
+ 16 2 = 0 ⇒ (25D2 − 40DD0 + 16D02 )z = 0
∂x
∂x∂y
∂y
The auxiliary equation is
25m2 − 40m + 16 = 0
4 4
m= ,
(repeated roots)
5 5 4
4
CF = φ1 y + x + xφ2 y + x = f1 (5y + 4x) + xf2 (5y + 4x)
5
5
F (x, y) = 0 ⇒ P I = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
44 / 55
Homogeneous Linear PDEs of Higher Order
Complementary function
Example: Solve 25r − 40s + 16t = 0
Solution: The equation can be written as
∂ 2z
∂ 2z
∂ 2z
25 2 − 40
+ 16 2 = 0 ⇒ (25D2 − 40DD0 + 16D02 )z = 0
∂x
∂x∂y
∂y
The auxiliary equation is
25m2 − 40m + 16 = 0
4 4
m= ,
(repeated roots)
5 5 4
4
CF = φ1 y + x + xφ2 y + x = f1 (5y + 4x) + xf2 (5y + 4x)
5
5
F (x, y) = 0 ⇒ P I = 0
Hence, the complete solution is
(P P Savani University)
Partial Differential Equation
March 16, 2022
44 / 55
Homogeneous Linear PDEs of Higher Order
Complementary function
Example: Solve 25r − 40s + 16t = 0
Solution: The equation can be written as
∂ 2z
∂ 2z
∂ 2z
25 2 − 40
+ 16 2 = 0 ⇒ (25D2 − 40DD0 + 16D02 )z = 0
∂x
∂x∂y
∂y
The auxiliary equation is
25m2 − 40m + 16 = 0
4 4
m= ,
(repeated roots)
5 5 4
4
CF = φ1 y + x + xφ2 y + x = f1 (5y + 4x) + xf2 (5y + 4x)
5
5
F (x, y) = 0 ⇒ P I = 0
Hence, the complete solution is
z = f1 (5y + 4x) + xf2 (5y + 4x)
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
44 / 55
Homogeneous Linear PDEs of Higher Order
Complementary function
Examples for Practice
Solve the following PDEs:
∂ 2z
∂ 2z
∂ 2z
1
−
−
6
=0
∂x2 ∂x∂y
∂y 2
2
4r + 12s + 9t = 0
3
r − 4s + 4t = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
45 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Particular Integral
Let the partial differential equation be f (D, D0 )z = F (x, y)
1
Particular Integral P.I. =
F (x, y)
f (D, D0 )
(P P Savani University)
Partial Differential Equation
March 16, 2022
46 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Particular Integral
Let the partial differential equation be f (D, D0 )z = F (x, y)
1
Particular Integral P.I. =
F (x, y)
f (D, D0 )
1
1
When F (x, y) = eax+by , P.I. =
eax+by [ Put D = a and D0 = b ]
f (D, D0 )
(P P Savani University)
Partial Differential Equation
March 16, 2022
46 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Particular Integral
Let the partial differential equation be f (D, D0 )z = F (x, y)
1
Particular Integral P.I. =
F (x, y)
f (D, D0 )
1
1
When F (x, y) = eax+by , P.I. =
eax+by [ Put D = a and D0 = b ]
f (D, D0 )
2
When F (x, y) = sin(ax + by) or cos(ax + by),
1
P.I. =
sin(ax + by) or cos(ax + by)
f (D2 , DD0 , D02 )
[Put D2 = −a2 , DD0 = −ab, D02 = −b2 ]
(P P Savani University)
Partial Differential Equation
March 16, 2022
46 / 55
Homogeneous Linear PDEs of Higher Order
3
Particular Integral
1
xm y n = [f (D, D0 )]−1 xm y n
f (D, D0 )
using binomial expansion according to the following rules:
When F (x, y) = xm y n , P.I. =
Expand [f (D, D0 )]−1
i
ii
If m > n, expand in power of
If n > m, expand in power of
(P P Savani University)
D0
D
D
D0
Partial Differential Equation
March 16, 2022
47 / 55
Homogeneous Linear PDEs of Higher Order
3
1
xm y n = [f (D, D0 )]−1 xm y n
f (D, D0 )
using binomial expansion according to the following rules:
When F (x, y) = xm y n , P.I. =
Expand [f (D, D0 )]−1
i
ii
4
Particular Integral
If m > n, expand in power of
If n > m, expand in power of
D0
D
D
D0
When F (x, y) is any function of x and y P.I. =
(P P Savani University)
Partial Differential Equation
1
F (x, y)
f (D, D0 )
March 16, 2022
47 / 55
Homogeneous Linear PDEs of Higher Order
3
1
xm y n = [f (D, D0 )]−1 xm y n
f (D, D0 )
using binomial expansion according to the following rules:
When F (x, y) = xm y n , P.I. =
Expand [f (D, D0 )]−1
i
ii
4
Particular Integral
If m > n, expand in power of
If n > m, expand in power of
D0
D
D
D0
When F (x, y) is any function of x and y P.I. =
1
F (x, y)
f (D, D0 )
1
using
f (D, D0 )
the partial fractions method. Operate each part on F (x, y) considering
Z
1
F (x, y) = F (x, c − mx)dx,
D − mD0
Express f (D, D0 ) in linear factors of D and separate each factor of
where c is replaced by y + mx after integration.
(P P Savani University)
Partial Differential Equation
March 16, 2022
47 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D3 − 3D2 D0 + 4D03 )z = ex+2y
(P P Savani University)
Partial Differential Equation
March 16, 2022
48 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D3 − 3D2 D0 + 4D03 )z = ex+2y
Solution: The auxiliary equation is
(P P Savani University)
Partial Differential Equation
March 16, 2022
48 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D3 − 3D2 D0 + 4D03 )z = ex+2y
Solution: The auxiliary equation is
m3 − 3m2 + 4 = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
48 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D3 − 3D2 D0 + 4D03 )z = ex+2y
Solution: The auxiliary equation is
m3 − 3m2 + 4 = 0
(m + 1)(m − 2)2 = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
48 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D3 − 3D2 D0 + 4D03 )z = ex+2y
Solution: The auxiliary equation is
m3 − 3m2 + 4 = 0
(m + 1)(m − 2)2 = 0
m = −1, 2, 2
(P P Savani University)
Partial Differential Equation
March 16, 2022
48 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D3 − 3D2 D0 + 4D03 )z = ex+2y
Solution: The auxiliary equation is
m3 − 3m2 + 4 = 0
(m + 1)(m − 2)2 = 0
m = −1, 2, 2
CF = f1 (y − x) + f2 (y + 2x) + xf3 (y + 2x)
(P P Savani University)
Partial Differential Equation
March 16, 2022
48 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D3 − 3D2 D0 + 4D03 )z = ex+2y
Solution: The auxiliary equation is
m3 − 3m2 + 4 = 0
(m + 1)(m − 2)2 = 0
m = −1, 2, 2
CF = f1 (y − x) + f2 (y + 2x) + xf3 (y + 2x)
1
PI = 3
ex+2y [Replace D by a = 1 and D0 by b = 2]
D − 3DD0 + 4D03
(P P Savani University)
Partial Differential Equation
March 16, 2022
48 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D3 − 3D2 D0 + 4D03 )z = ex+2y
Solution: The auxiliary equation is
m3 − 3m2 + 4 = 0
(m + 1)(m − 2)2 = 0
m = −1, 2, 2
CF = f1 (y − x) + f2 (y + 2x) + xf3 (y + 2x)
1
PI = 3
ex+2y [Replace D by a = 1 and D0 by b = 2]
D − 3DD0 + 4D03
ex+2y
ex+2y
=
=
3
1 − 3(1)(2) + 4(2)
27
(P P Savani University)
Partial Differential Equation
March 16, 2022
48 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D3 − 3D2 D0 + 4D03 )z = ex+2y
Solution: The auxiliary equation is
m3 − 3m2 + 4 = 0
(m + 1)(m − 2)2 = 0
m = −1, 2, 2
CF = f1 (y − x) + f2 (y + 2x) + xf3 (y + 2x)
1
PI = 3
ex+2y [Replace D by a = 1 and D0 by b = 2]
D − 3DD0 + 4D03
ex+2y
ex+2y
=
=
3
1 − 3(1)(2) + 4(2)
27
Hence, the complete solution is
(P P Savani University)
Partial Differential Equation
March 16, 2022
48 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D3 − 3D2 D0 + 4D03 )z = ex+2y
Solution: The auxiliary equation is
m3 − 3m2 + 4 = 0
(m + 1)(m − 2)2 = 0
m = −1, 2, 2
CF = f1 (y − x) + f2 (y + 2x) + xf3 (y + 2x)
1
PI = 3
ex+2y [Replace D by a = 1 and D0 by b = 2]
D − 3DD0 + 4D03
ex+2y
ex+2y
=
=
3
1 − 3(1)(2) + 4(2)
27
Hence, the complete solution is
ex+2y
z = f1 (y − x) + f2 (y + 2x) + xf3 (y + 2x) +
27
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
48 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D2 − 4DD0 + 4D02 )z = cos(x − 2y)
(P P Savani University)
Partial Differential Equation
March 16, 2022
49 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D2 − 4DD0 + 4D02 )z = cos(x − 2y)
Solution: The auxiliary equation is
(P P Savani University)
Partial Differential Equation
March 16, 2022
49 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D2 − 4DD0 + 4D02 )z = cos(x − 2y)
Solution: The auxiliary equation is
m2 − 4m + 4 = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
49 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D2 − 4DD0 + 4D02 )z = cos(x − 2y)
Solution: The auxiliary equation is
m2 − 4m + 4 = 0
m = 2, 2
(P P Savani University)
Partial Differential Equation
March 16, 2022
49 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D2 − 4DD0 + 4D02 )z = cos(x − 2y)
Solution: The auxiliary equation is
m2 − 4m + 4 = 0
m = 2, 2
CF = f1 (y + 2x) + xf2 (y + 2x)
(P P Savani University)
Partial Differential Equation
March 16, 2022
49 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D2 − 4DD0 + 4D02 )z = cos(x − 2y)
Solution: The auxiliary equation is
m2 − 4m + 4 = 0
m = 2, 2
CF = f1 (y + 2x) + xf2 (y + 2x)
1
PI = 2
cos(x − 2y)
D − 4DD0 + 4D02
[Replace D2 by − a2 = −1, DD0 by − ab = 2 and D02 by − b2 = −4]
(P P Savani University)
Partial Differential Equation
March 16, 2022
49 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D2 − 4DD0 + 4D02 )z = cos(x − 2y)
Solution: The auxiliary equation is
m2 − 4m + 4 = 0
m = 2, 2
CF = f1 (y + 2x) + xf2 (y + 2x)
1
PI = 2
cos(x − 2y)
D − 4DD0 + 4D02
[Replace D2 by − a2 = −1, DD0 by − ab = 2 and D02 by − b2 = −4]
PI =
1
cos(x − 2y)
cos(x − 2y) = −
(−1) − 4(2) + 4(−4)
25
(P P Savani University)
Partial Differential Equation
March 16, 2022
49 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D2 − 4DD0 + 4D02 )z = cos(x − 2y)
Solution: The auxiliary equation is
m2 − 4m + 4 = 0
m = 2, 2
CF = f1 (y + 2x) + xf2 (y + 2x)
1
PI = 2
cos(x − 2y)
D − 4DD0 + 4D02
[Replace D2 by − a2 = −1, DD0 by − ab = 2 and D02 by − b2 = −4]
1
cos(x − 2y)
cos(x − 2y) = −
(−1) − 4(2) + 4(−4)
25
Hence, the complete solution is
PI =
(P P Savani University)
Partial Differential Equation
March 16, 2022
49 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D2 − 4DD0 + 4D02 )z = cos(x − 2y)
Solution: The auxiliary equation is
m2 − 4m + 4 = 0
m = 2, 2
CF = f1 (y + 2x) + xf2 (y + 2x)
1
PI = 2
cos(x − 2y)
D − 4DD0 + 4D02
[Replace D2 by − a2 = −1, DD0 by − ab = 2 and D02 by − b2 = −4]
1
cos(x − 2y)
cos(x − 2y) = −
(−1) − 4(2) + 4(−4)
25
Hence, the complete solution is
cos(x − 2y)
z = f1 (y + 2x) + xf2 (y + 2x) −
25
PI =
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
49 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D2 − 2DD0 )z = x3 y + e5x
(P P Savani University)
Partial Differential Equation
March 16, 2022
50 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D2 − 2DD0 )z = x3 y + e5x
Solution: The auxiliary equation is
(P P Savani University)
Partial Differential Equation
March 16, 2022
50 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D2 − 2DD0 )z = x3 y + e5x
Solution: The auxiliary equation is
m2 − 2m = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
50 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D2 − 2DD0 )z = x3 y + e5x
Solution: The auxiliary equation is
m2 − 2m = 0
m = 0, 2
(P P Savani University)
Partial Differential Equation
March 16, 2022
50 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D2 − 2DD0 )z = x3 y + e5x
Solution: The auxiliary equation is
m2 − 2m = 0
m = 0, 2
CF = f1 (y) + f2 (y + 2x)
(P P Savani University)
Partial Differential Equation
March 16, 2022
50 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D2 − 2DD0 )z = x3 y + e5x
Solution: The auxiliary equation is
m2 − 2m = 0
m = 0, 2
CF = f1 (y) + f2 (y + 2x)
PI1 =
(P P Savani University)
x3 y
D2 − 2DD0
Partial Differential Equation
March 16, 2022
50 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D2 − 2DD0 )z = x3 y + e5x
Solution: The auxiliary equation is
m2 − 2m = 0
m = 0, 2
CF = f1 (y) + f2 (y + 2x)
x3 y
D2 − 2DD0
1
x3 y
= 2
2D0
D 1− D
PI1 =
(P P Savani University)
Partial Differential Equation
March 16, 2022
50 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D2 − 2DD0 )z = x3 y + e5x
Solution: The auxiliary equation is
m2 − 2m = 0
m = 0, 2
CF = f1 (y) + f2 (y + 2x)
x3 y
D2 − 2DD0
1
x3 y
= 2
2D0
D 1− D
−1
1
2D0
= 2 1−
(x3 y)
D
D
PI1 =
(P P Savani University)
Partial Differential Equation
March 16, 2022
50 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Example: Solve (D2 − 2DD0 )z = x3 y + e5x
Solution: The auxiliary equation is
m2 − 2m = 0
m = 0, 2
CF = f1 (y) + f2 (y + 2x)
x3 y
D2 − 2DD0
1
x3 y
= 2
2D0
D 1− D
−1
1
2D0
= 2 1−
(x3 y)
D
D
1
2D0 4D02 8D03
= 2 1+
+
+
+ · · · (x3 y)
D
D
D2
D3
PI1 =
(P P Savani University)
Partial Differential Equation
March 16, 2022
50 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
∂
∂
where D0 =
&D=
∂y
∂x
By simplification,
(P P Savani University)
Partial Differential Equation
March 16, 2022
51 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
∂
∂
where D0 =
&D=
∂y
∂x
By simplification,
1
2
PI1 = 2 x3 y + 3 x3
D
D
(P P Savani University)
Partial Differential Equation
March 16, 2022
51 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
∂
∂
where D0 =
&D=
∂y
∂x
By simplification,
1
2
PI1 = 2 x3 y + 3 x3
D
D
x5 y x6
=
+
20
60
(P P Savani University)
Partial Differential Equation
March 16, 2022
51 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
∂
∂
where D0 =
&D=
∂y
∂x
By simplification,
1
2
PI1 = 2 x3 y + 3 x3
D
D
x5 y x6
=
+
20
60
5x
e
Now, PI2 = 2
[Replace D by a = 5 and D0 by b = 0]
D − 2DD0
(P P Savani University)
Partial Differential Equation
March 16, 2022
51 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
∂
∂
where D0 =
&D=
∂y
∂x
By simplification,
1
2
PI1 = 2 x3 y + 3 x3
D
D
x5 y x6
=
+
20
60
5x
e
Now, PI2 = 2
[Replace D by a = 5 and D0 by b = 0]
D − 2DD0
e5x
=
25
(P P Savani University)
Partial Differential Equation
March 16, 2022
51 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
∂
∂
where D0 =
&D=
∂y
∂x
By simplification,
1
2
PI1 = 2 x3 y + 3 x3
D
D
x5 y x6
=
+
20
60
5x
e
Now, PI2 = 2
[Replace D by a = 5 and D0 by b = 0]
D − 2DD0
e5x
=
25
x5 y x6 e5x
PI = PI1 + PI2 =
+
+
20
60
25
(P P Savani University)
Partial Differential Equation
March 16, 2022
51 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
∂
∂
where D0 =
&D=
∂y
∂x
By simplification,
1
2
PI1 = 2 x3 y + 3 x3
D
D
x5 y x6
=
+
20
60
5x
e
Now, PI2 = 2
[Replace D by a = 5 and D0 by b = 0]
D − 2DD0
e5x
=
25
x5 y x6 e5x
PI = PI1 + PI2 =
+
+
20
60
25
Hence, the complete solution is
(P P Savani University)
Partial Differential Equation
March 16, 2022
51 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
∂
∂
where D0 =
&D=
∂y
∂x
By simplification,
1
2
PI1 = 2 x3 y + 3 x3
D
D
x5 y x6
=
+
20
60
5x
e
Now, PI2 = 2
[Replace D by a = 5 and D0 by b = 0]
D − 2DD0
e5x
=
25
x5 y x6 e5x
PI = PI1 + PI2 =
+
+
20
60
25
Hence, the complete solution is
x5 y x6 e5x
z = f1 (y) + f2 (y + 2x) +
+
+
20
60
25
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
51 / 55
Homogeneous Linear PDEs of Higher Order
Example: Solve 4
∂2z
∂x2
(P P Savani University)
−4
∂2z
∂x∂y
+
∂2z
∂y 2
Particular Integral
= 16 log(x + 2y)
Partial Differential Equation
March 16, 2022
52 / 55
Homogeneous Linear PDEs of Higher Order
∂2z
∂2z
Particular Integral
∂2z
−4
+
= 16 log(x + 2y)
∂x2
∂x∂y ∂y 2
Solution: The equation can be written as 4D2 − 4DD0 + D02 = 16 log(x + 2y)
Example: Solve 4
(P P Savani University)
Partial Differential Equation
March 16, 2022
52 / 55
Homogeneous Linear PDEs of Higher Order
∂2z
∂2z
Particular Integral
∂2z
−4
+
= 16 log(x + 2y)
∂x2
∂x∂y ∂y 2
Solution: The equation can be written as 4D2 − 4DD0 + D02 = 16 log(x + 2y)
Example: Solve 4
The auxiliary equation is
(P P Savani University)
Partial Differential Equation
March 16, 2022
52 / 55
Homogeneous Linear PDEs of Higher Order
∂2z
∂2z
Particular Integral
∂2z
−4
+
= 16 log(x + 2y)
∂x2
∂x∂y ∂y 2
Solution: The equation can be written as 4D2 − 4DD0 + D02 = 16 log(x + 2y)
Example: Solve 4
The auxiliary equation is
4m2 − 4m + 1 = 0
(P P Savani University)
Partial Differential Equation
March 16, 2022
52 / 55
Homogeneous Linear PDEs of Higher Order
∂2z
∂2z
Particular Integral
∂2z
−4
+
= 16 log(x + 2y)
∂x2
∂x∂y ∂y 2
Solution: The equation can be written as 4D2 − 4DD0 + D02 = 16 log(x + 2y)
Example: Solve 4
The auxiliary equation is
4m2 − 4m + 1 = 0
1 1
m= ,
2 2
(P P Savani University)
Partial Differential Equation
March 16, 2022
52 / 55
Homogeneous Linear PDEs of Higher Order
∂2z
∂2z
Particular Integral
∂2z
−4
+
= 16 log(x + 2y)
∂x2
∂x∂y ∂y 2
Solution: The equation can be written as 4D2 − 4DD0 + D02 = 16 log(x + 2y)
Example: Solve 4
The auxiliary equation is
4m2 − 4m + 1 = 0
1 1
m= ,
2 2
CF = f1
(P P Savani University)
1
1
y + x + xf2 y + x
2
2
Partial Differential Equation
March 16, 2022
52 / 55
Homogeneous Linear PDEs of Higher Order
∂2z
∂2z
Particular Integral
∂2z
−4
+
= 16 log(x + 2y)
∂x2
∂x∂y ∂y 2
Solution: The equation can be written as 4D2 − 4DD0 + D02 = 16 log(x + 2y)
Example: Solve 4
The auxiliary equation is
4m2 − 4m + 1 = 0
1 1
m= ,
2 2
1
1
CF = f1 y + x + xf2 y + x
2
2
1
PI =
16 log(x + 2y)
(2D − D0 )2
(P P Savani University)
Partial Differential Equation
March 16, 2022
52 / 55
Homogeneous Linear PDEs of Higher Order
∂2z
∂2z
Particular Integral
∂2z
−4
+
= 16 log(x + 2y)
∂x2
∂x∂y ∂y 2
Solution: The equation can be written as 4D2 − 4DD0 + D02 = 16 log(x + 2y)
Example: Solve 4
The auxiliary equation is
4m2 − 4m + 1 = 0
1 1
m= ,
2 2
1
1
CF = f1 y + x + xf2 y + x
2
2
1
PI =
16 log(x + 2y)
(2D − D0 )2
"
#
1
1
log(x + 2y)
=4
D − 21 D0
D − 12 D0
(P P Savani University)
Partial Differential Equation
March 16, 2022
52 / 55
Homogeneous Linear PDEs of Higher Order
PI = 4
1
D − 12 D0
(P P Savani University)
Z
Particular Integral
n
x o
log x + 2 c −
dx
2
Partial Differential Equation
h
∵ y = c − mx = c −
xi
2
March 16, 2022
53 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Z
n
x o
1
log x + 2 c −
PI = 4
dx
2
D − 12 D0
Z
1
log(2c) 1dx
=4
D − 12 D0
(P P Savani University)
Partial Differential Equation
h
∵ y = c − mx = c −
xi
2
March 16, 2022
53 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Z
n
x o
1
log x + 2 c −
PI = 4
dx
2
D − 12 D0
Z
1
log(2c) 1dx
=4
D − 12 D0
1
[x log(x + 2y)]
=4
D − 12 D0
(P P Savani University)
Partial Differential Equation
h
∵ y = c − mx = c −
h
c = y + mx = y +
xi
2
xi
2
March 16, 2022
53 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Z
n
x o
1
log x + 2 c −
PI = 4
dx
2
D − 12 D0
Z
1
log(2c) 1dx
=4
D − 12 D0
1
[x log(x + 2y)]
=4
D − 12 D0
Z n
h
x io
=4
x log x + 2 c −
dx
2
(P P Savani University)
Partial Differential Equation
h
∵ y = c − mx = c −
h
c = y + mx = y +
xi
2
xi
2
March 16, 2022
53 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Z
n
x o
1
log x + 2 c −
PI = 4
dx
2
D − 12 D0
Z
1
log(2c) 1dx
=4
D − 12 D0
1
[x log(x + 2y)]
=4
D − 12 D0
Z n
h
x io
=4
x log x + 2 c −
dx
2
Z
= 4 log 2c xdx
h
∵ y = c − mx = c −
h
c = y + mx = y +
xi
2
xi
2
= 2x2 log(x + 2y)
(P P Savani University)
Partial Differential Equation
March 16, 2022
53 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Z
n
x o
1
log x + 2 c −
PI = 4
dx
2
D − 12 D0
Z
1
log(2c) 1dx
=4
D − 12 D0
1
[x log(x + 2y)]
=4
D − 12 D0
Z n
h
x io
=4
x log x + 2 c −
dx
2
Z
= 4 log 2c xdx
h
∵ y = c − mx = c −
h
c = y + mx = y +
xi
2
xi
2
= 2x2 log(x + 2y)
Hence, the complete solution is
(P P Savani University)
Partial Differential Equation
March 16, 2022
53 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Z
n
x o
1
log x + 2 c −
PI = 4
dx
2
D − 12 D0
Z
1
log(2c) 1dx
=4
D − 12 D0
1
[x log(x + 2y)]
=4
D − 12 D0
Z n
h
x io
=4
x log x + 2 c −
dx
2
Z
= 4 log 2c xdx
h
∵ y = c − mx = c −
h
c = y + mx = y +
xi
2
xi
2
= 2x2 log(x + 2y)
Hence, the complete solution
is 1
1
z = f1 y + x + xf2 y + x + 2x2 log(x + 2y)
2
2
J
(P P Savani University)
Partial Differential Equation
March 16, 2022
53 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
Examples for Practice
Solve the following PDEs:
1
(D2 + 10DD0 + 25D02 )z = e3x+2y
2
(D2 − 2DD0 + D02 )z = ex+2y + x3
3
(D2 − 2DD0 )z = sin x cos 2y
4
(D2 − 2DD0 + D02 )z = tan(y + x)
(P P Savani University)
Partial Differential Equation
March 16, 2022
54 / 55
Homogeneous Linear PDEs of Higher Order
Particular Integral
τ ~αnk Ψφu !!
(P P Savani University)
Partial Differential Equation
March 16, 2022
55 / 55