Outline
1
Motivation
2
Basic Ideas and Terminology
3
Solutions of Differential Equations
4
Initial-Value Problem
5
Separable Differential Equations
6
Some Simple Population Models
7
First-Order Linear Differential Equations
8
Modeling Problems Using First-order Linear Differential Equations
9
Change of Variables
10
Exact Differential Equations
11
Numerical Solution to First-order Differential Equations
MATH2030 Differential Equations (AY 2023-24)
104 / 177
Change of Variables
In this section, we consider two further types of differential equations
which can be solved by using a change of variables to reduce them
to one of the types we know how to solve.
Definition
A function f (x, y ) is said to be homogeneous of degree zero if
f (tx, ty ) = f (x, y )
for all positive values of t for which (tx, ty ) is in the domain of f .
Equivalently, f is homogeneous of degree zero if it is invariant under a
re-scaling of the variables x and y .
MATH2030 Differential Equations (AY 2023-24)
105 / 177
Examples
1
The simplest non-constant functions which are homogeneous of
y
x
degree zero are f (x, y ) = and f (x, y ) = .
x
y
y
Indeed, for say f (x, y ) = we have
x
f (tx, ty ) =
2
If f (x, y ) =
ty
y
= = f (x, y ).
tx
x
x2 − y2
, then
2xy + y 2
f (tx, ty ) =
t 2 (x 2 − y 2 )
= f (x, y ),
t 2 (2xy + y 2 )
so that f is homogeneous of degree zero.
MATH2030 Differential Equations (AY 2023-24)
106 / 177
Note that in the second example, dividing both the numerator and
denominator by x 2 , we have
x2 − y2
=
f (x, y ) =
2xy + y 2
y2
x2
−
2
x
x2
2
2xy
+ yx 2
x2
=
2
1−
y
x
y 2
x
2 .
+ yx
This means that f can be considered to depend on the single variable
y
V = .
x
MATH2030 Differential Equations (AY 2023-24)
107 / 177
Actually, we have the following result that establishes the basic
property of all functions that are homogeneous of degree zero.
Theorem
A function f (x, y ) is homogeneous of degree zero if and only if it
y
x
depends on either or only.
x
y
MATH2030 Differential Equations (AY 2023-24)
108 / 177
Homogeneous first-order differential equations
We now consider differential equations that satisfy the following
definition.
Definition
If f (x, y ) is homogeneous of degree zero, then the differential equation
dy
= f (x, y )
dx
(1)
is called a homogeneous first-order differential equation.
MATH2030 Differential Equations (AY 2023-24)
109 / 177
In general, a homogeneous first-order differential equation cannot be
solved directly.
However, we can solve it, by the theorem above, using a change of
variables: instead of using variables x and y , we can use the variables
y
x and V , where V = , that is y = xV (x).
x
MATH2030 Differential Equations (AY 2023-24)
110 / 177
Theorem
The change of variables y = xV (x) reduces a homogeneous
dy
first-order differential equation
= f (x, y ) to the separable equation
dx
1
1
dV = dx,
F (V ) − V
x
where F is an appropriate function.
Proof
Exrecise.
MATH2030 Differential Equations (AY 2023-24)
111 / 177
Remark
The separable equation said above can be integrated to obtain a
relationship between V and x. Then the solution to the given
y
differential equation can be obtained by substituting for V in this
x
relationship.
Examples
Find the general solution to
dy
4x + y
=
.
dx
x − 4y
Solution
Since f (x, y ) = 4x+y
x−4y is homogeneous of degree zero, the given
equation is a first-order homogeneous differential equation.
MATH2030 Differential Equations (AY 2023-24)
112 / 177
Substituting y = xV into the equation yields
4+V
d
(xV ) =
,
dx
1 − 4V
that is
x
dV
4+V
+V =
,
dx
1 − 4V
or equivalently,
x
4(1 + V 2 )
dV
=
.
dx
1 − 4V
Separating the variables gives
1 − 4V
1
dV = dx.
2
x
4(1 + V )
MATH2030 Differential Equations (AY 2023-24)
113 / 177
We write this as
1
V
1
−
dV = dx.
2
2
x
4(1 + V ) 1 + V
Now we integrate both side directly. For this the following integral
formulas are used:
Z
Z
dx
x dx
1
(1)
= arctan x + C, and (2)
= log(1 + x 2 ) + C.
2
2
2
1+x
1+x
The first integral is standard: the change of variables x = tan t works
well.
For the second integral: putting 1 + x 2 = t we have dt = 2x dx, and
hence
Z
Z
x dx
1
dt
1
1
=
= log |t| + C = log(1 + x 2 ) + C.
2
2
t
2
2
1+x
MATH2030 Differential Equations (AY 2023-24)
114 / 177
Thus we obtain
1
1
arctan V − log(1 + V 2 ) = log |x| + C.
4
2
y
and multiplying through by 2 yields
x
2
y 1
x + y2
arctan
− log
= log(x 2 ) + C1 ,
2
x
x2
Substituting back V =
where C1 =2C. 2
2
Moving log x x+y
to the right-hand side yields the answer
2
y 1
arctan
− log(x 2 + y 2 ) = C,
2
x
where C is an arbitrary constant.
MATH2030 Differential Equations (AY 2023-24)
115 / 177
Remark
The solution to the example above is more easily expressed in terms
of polar coordinates.

x = r cos θ
Indeed, note that
y = r sin θ
obtain a logarithmic spiral:

p
r = x 2 + y 2
⇐⇒
θ = arctan y ,
we then
x
θ
r = Ce 4
MATH2030 Differential Equations (AY 2023-24)
116 / 177
Bernoulli equations
We now consider a special type of nonlinear differential equation that
can be reduced to a linear equation by a change of variables.
Definition
A differential equation that can be written in the form
dy
+ p(x)y = q(x)y α ,
dx
(2)
where α is a real constant, is called a Bernoulli equation.
If α = 0 or α = 1, Equation (2) is linear, but otherwise it is nonlinear.
In general, we can reduce Equation (2) to a linear equation as follows.
MATH2030 Differential Equations (AY 2023-24)
117 / 177
Divide Equation (2) by y α to obtain
y −α
dy
+ y 1−α p(x) = q(x).
dx
(3)
Make the change of variables u(x) = y 1−α to get
dy
du
= (1 − α)y −α .
dx
dx
Thus
y −α
dy
1 du
=
.
dx
1 − α dx
MATH2030 Differential Equations (AY 2023-24)
118 / 177
Now substituting into Equation (3) the expressions y 1−α = u(x) and
dy
1 du
y −α
=
, we have the linear differential equation
dx
1 − α dx
1 du
+ p(x)u = q(x),
1 − α dx
or in the standard form
du
+ (1 − α)p(x)u = (1 − α)q(x).
dx
MATH2030 Differential Equations (AY 2023-24)
119 / 177
Examples
Solve
2
dy
3
12y 3
+ y=√
, x > 0.
dx
x
1 + x2
Solution
The differential equation is Bernoulli equation with α = 23 .
2
Dividing both sides of the equation by y α = y 3 we obtain
2
y−3
dy
3 1
12
+ y3 = √
.
dx
x
1 + x2
(4)
We make the change of variables
1
u = y 1−α = y 3 .
MATH2030 Differential Equations (AY 2023-24)
120 / 177
Then we get
1 2 dy
du
= y−3
.
dx
3
dx
Now substitute the last two expressions into Equation (4) yields
3
du
3
12
+ u=√
,
dx
x
1 + x2
or in the standard form
du
1
4
+ u=√
.
dx
x
1 + x2
(5)
This is a first-order linear differential equation, whose integrating
factor is
MATH2030 Differential Equations (AY 2023-24)
121 / 177
I(x) = e
R
1
dx
x
= elog x (because x > 0) = x.
So Equation (5) can be written as
d
4x
(xu) = √
.
dx
1 + x2
Integrating this equation, we obtain
p
xu(x) = 4 1 + x 2 + C.
Here we use the integral formula
Z
p
x dx
√
= 1 + x 2 + C,
1 + x2
which is computed as follows: putting 1 + x 2 = t implies x dx =
MATH2030 Differential Equations (AY 2023-24)
dt
2;
122 / 177
then
Z
x dx
√
=
1 + x2
p
√
dt
√ = t + C = 1 + x 2 + C.
2 t
Z
Returning back to the obtained result
p
xu(x) = 4 1 + x 2 + C,
we have
1 p
4 1 + x2 + C ,
x
and the solution to the original differential equation is
u(x) =
1
y3 =
1 p
4 1 + x2 + C ,
x
where C is an arbitrary constant.
MATH2030 Differential Equations (AY 2023-24)
123 / 177
Outline
1
Motivation
2
Basic Ideas and Terminology
3
Solutions of Differential Equations
4
Initial-Value Problem
5
Separable Differential Equations
6
Some Simple Population Models
7
First-Order Linear Differential Equations
8
Modeling Problems Using First-order Linear Differential Equations
9
Change of Variables
10
Exact Differential Equations
11
Numerical Solution to First-order Differential Equations
MATH2030 Differential Equations (AY 2023-24)
124 / 177
Exact Differential Equations
The technique presented in this section is for first-order differential
equations written in a differential form
M(x, y ) dx + N(x, y ) dy = 0.
Recall from a calculus course that if u = u(x, y ) is a function of two
variables x and y , then its differential du is defined by
du =
∂u
∂u
dx +
dy .
∂x
∂y
MATH2030 Differential Equations (AY 2023-24)
125 / 177
Example
Solve the equation
2x sin y dx + x 2 cos y dy = 0.
(1)
Solution
This equation is separable, but we will use a different technique to
solve it.
We notice that 2x sin y dx + x 2 cos y dy = d(x 2 sin y ). Then Equation
(1) can be written as d(x 2 sin y ) = 0, which implies that x 2 sin y is
constant, denoted by C. Thus the general solution to Equation (1) is
sin y =
C
,
x2
where C is an arbitrary constant.
MATH2030 Differential Equations (AY 2023-24)
126 / 177
Obviously, in the example above, it was able to write the given
differential equation in the form du(x, y ) = 0, and hence obtain its
solution.
However, we cannot always do it. Indeed, we see that the differential
equation
M(x, y ) dx + N(x, y ) dy = 0
can be written as du = 0 if and only if
M=
∂u
∂x
and N =
∂u
∂y
for some function u.
This leads to the following definition:
MATH2030 Differential Equations (AY 2023-24)
127 / 177
Definition
The differential equation
M(x, y ) dx + N(x, y ) dy = 0
(2)
is said to be exact in a domain D of the xy -plane if there exists a
function u(x, y ) such that
∂u
∂u
= M,
= N,
∂x
∂y
(3)
for all (x, y ) ∈ D.
Any function u satisfying (3) is called a potential function for the
Equation (2).
MATH2030 Differential Equations (AY 2023-24)
128 / 177
If such a function does exist, then Equation (2) can be written as
M(x, y ) dx + N(x, y ) dy = 0
∂u
∂u
⇐⇒
dx +
dy = 0
∂x
∂y
⇐⇒ du = 0.
This is a reason why such a differential equation is called an exact
differential equation.
For the example above, the potential function is
u(x, y ) = x 2 sin y .
MATH2030 Differential Equations (AY 2023-24)
129 / 177
The following result shows that if a differential equation is exact and if a
potential function u can be found, then we can write down its solution
immediately.
Theorem
The general solution to an exact differential equation
M(x, y ) dx + N(x, y ) dy = 0
is defined implicitly by
u(x, y ) = C,
(4)
where u(x, y ) satisfies (3) and C is an arbitrary constant.
MATH2030 Differential Equations (AY 2023-24)
130 / 177
Proof
Exrecise.
MATH2030 Differential Equations (AY 2023-24)
131 / 177
There naturally arise 2 questions:
1
How to verify if a given differential equation is exact?
2
How to find a potential function for an exact equation?
The answer is given in the following theorem:
Theorem (Test for Exactness)
Let M, N, and their first partial derivatives My and Nx , be continuous
in a simply connected domain D of the xy -plane. Then the differential
equation
M(x, y )dx + N(x, y )dy = 0
is exact for all (x, y ) in D if and only if
∂M
∂N
=
, that is My = Nx .
∂y
∂x
MATH2030 Differential Equations (AY 2023-24)
132 / 177
Here a simply connected domain D is a domain such that every
simple closed contour within it encloses only points of D.
The set of points interior to a simple closed contour is an example.
The annular domain between two concentric circles is not simply
connected.
A domain that is not simply connected is said to be multiply connected.
Remark
Since there is no need to memorize the final result for u, we omit the
details of the proof of the theorem above.
For a particular problem one can construct an appropriate potential
functions from first principles.
MATH2030 Differential Equations (AY 2023-24)
133 / 177
Examples
Determine whether the given differential equation is exact.
x
y dy
y 3 ) dy
1
[1 + log(xy )] dx +
= 0.
2
x 2 y dx − (xy 2 +
= 0.
Solution
1

M(x, y ) = [1 + log(xy )] =⇒ M =
y
We have
N(x, y ) = x =⇒ N = 1 .
y
2
x
1
y
y
So My = Nx and the differential equation is exact.

M(x, y ) = x 2 y =⇒ M = x 2
y
We have
N(x, y ) = −(xy 2 + y 3 ) =⇒ N = −y 2 .
x
Thus My 6= Nx , and hence the differential equation is not exact.
MATH2030 Differential Equations (AY 2023-24)
134 / 177