ORDINARY DIFFERENTIAL EQUATIONS
AKEJU A.O
University of Ibadan
Department of Mathematics
1
Course Outline
1. Introduction
2. Derivation of Differential Equations
3. Techniques for Solving Order 1 Linear and Non-Linear
4. Techniques for Solving Order 2 Linear and Non-Linear
5. Techniques for Solving nth Order Linear Equations
6. Finite Difference/Difference Equations
7. Numerical Differentiation/Integration: Interpolations, Errors.
Reference Materials
1. Ordinary Differential Equations and Applications by E.O. Ayoola
2. Schaum’s Outline Series of Differential Equations by Frank Ayres JR
3. Elementary Differential Equations by Earl D. Rainville
4. Numerical Analysis by Ian Jacques and Collin Judd
5. Engineering Mathematics by K. A. Stroud
Chapter 1
Introduction
An ODE is an equation that involves the unknown dependent function and
their derivatives which involve coefficients that are functions of the independent variables.
Definition:ORDER
The order of a differential equation is the highest derivative present in the
differential equation
Definition : DEGREE
The degree of a differential equation is the highest power present in the equation
Definition:LINEAR DIFFERENTIAL EQUATIONS
A D.E is called linear if there is no product of the dependent variable (function) and it derivative,and neither the function or its derivative occurs to any
power other than the first power
Definition:NON LINEAR DIFFERENTIAL EQUATIONS
If the differential equation is not linear,it is called non linear differential equation
2
3
Definition: SOLUTION
A solution to a differential equation on the interval a ≤ x ≤ b is any function y(x) which satisfies the differential equation in question on the interval
a≤x≤b
Solution are often accompanied by interval and these intervals can impart some
important information about the solution
The solution of ODE are classified into (1)General Solution and (2)Particular
Solution
Definition:INITIAL CONDITION(S)
This is a condition or set of conditions on the solution that allow us to determine the particular solution and which solution that we are after.Initial
Condition are values of the solution and its derivative at specific points.The
number of I.C required for the solution of a certain differential equation will
depend on the order of the D.E
Definition:INITIAL VALUE PROBLEM
An initial value problem (IVP) is a differential equation along with an appropriate number of initial conditions
Examples are:
•
dy
= cosx
dx
order 1, linear equation, (y is dependent variable, x is independent variable)
4
•
d2 y
+ k2y = 0
2
dx
order 2, linear equation, (y is dependent variable, x is independent variable)
•
d3 x
dx
+
x
− 4xy = 0
dy 3
dy
order 3, linear equation, (x is dependent variable, y is independent variable)
•
(
dw
d2 w 3
)
−
xy
+w =0
dx2
dx
order 2, degree 3, non linear equation,(w is dependent variable, x is
independent variable)
When an equation involves one or more derivatives with respect to a particular variable, that variable is called an independent variable.
∂ 2v ∂ 2v
+
=0
∂x2 ∂y 2
PDE order 2, V is dependent, x and y are independent variable
Generally, the equation
F (x, y, y 0 , y 00 , . . . , y (n) ) = 0
(1.1)
5
is called an n th-order ODE. Equation (1.1) can be solved explicitly for y (n) in
terms of the other n + 1 variables x, y, y0, y 00 , . . . , to obtain
y (n) = f (x, y, y 0 , y 00 , . . . , y (n−1) )
(1.2)
ODE arises from geometric, physical, biological problems (e.t.c.)
Illustration
If the population of a country doubles in 50 years, in how many years will it
become triple under the assumption that the rate of increase is proportional
to the number of inhabitants?
Solution:
Let y denote the population at time t years and y0 the population at time
t = 0. Then
dy
dy
dy
∝ y =⇒
= ky =⇒
= kdt.
dt
dt
y
where k is constant.
Integrating both sides,we have
ln y = kt + ln c
y = cekt
At time
t = 0, y = y0 =⇒ y0 = ce0 =⇒ y0 = c =⇒ y = y0 ekt .
At time t = 50 years, y = 2y0 , 2y0 = y0 e50 k , 2 = e50k
6
when y = 3y0 ,
3 = ekt
so
350 = e50kt = (e50k )t = 2t
350 = 2t =⇒ t = 79years.
Exercises
State the type, linearity, order ,independent and dependent variables of the
following equations:
1.
d2 x
+ k2x = 0
2
dt
2.
(x2 + y 2 )dx + 2xydy = 0
3.
(
d3 w 2
dw
) − 2( )4 + yw = 0
3
dx
dx
4.
2
∂ 2w
2∂ w
=
a
∂t2
∂x2
5.
x(y 00 )3 + (y 0 )4 − y = 0
Chapter 2
Derivative of ODE
Just as a function can be obtained as a solution of any given differential equation, the reverse process can be carried out by obtaining an ODE from a given
function through the process of eliminating some constants that appear in such
functions.
Examples:
Form a DE by eliminating the constant(s) in the following equations:
1. y = Ax
Solution:
y0 =
dy
y
dy
= A =⇒
= y0 =
dx
dx
x
2. y = Ax + B
Solution:
y 0 = A, y 00 = 0
but
y0 = y −
B 00 xy 0 − y + B
,y =
x
x2
7
8
3. y = Ax2 + Bx + c
Solution:
y 0 = 2Ax + B
y 00 = 2A
y 000 = 0 [free or arbitrary constants]
4. y = A cos ax + B sin ax
Solution:
y 0 = −Aa sin ax + Ba cos ax
y 00 = −Aa2 cos ax − Ba2 sin ax = −a2 (A cos ax + B sin ax) = −a2 y
y 00 + a2 y = 0
5. y = x +
A
x
= x + Ax−1
Solution
y 0 = 1 − Ax−2
but
A
x
= y − x =⇒ A = x(y − x)
=
y 0 = 1 − x y−x
x2
x−y+x
x
=
2x−y
x
Exercise
Form a differential equation from these functions
• y = Ax2 + Bx
• y = −Asinx + Bcosx
• y = c1 e3x + c2 e2x + c3 ex
9
Note: A function with 1 arbitrary constant gives a 1st order equation. A
function with 2 arbitrary constants gives a 2nd order equation. A function
with n arbitrary constant gives a n-th order equation.
Chapter 3
Technique for Solving Order 1
Linear and Non Linear ODE
We shall classify this into five categories as follows;
• Separation of Variable
• Homogeneous type
• Exact type
• Non exact type
• Bernoulli equation
3.1
Separation of Variable
If a first order ODE of the form
dy
= f (x)g(y)
dx
can be written in the form
dy
= f (x)dx
g(y)
10
11
then the equation is a separable equation where f (x) and g(y) are given functions of the independent variable and the dependent variable respectively.
Examples
•
dy
= 2xy 2
dx
•
y −1
dy
(x + 1)−1
dx
•
(3y 2 + ey )
dy
= cosx
dx
•
(x + 1)
dy
= 2y
dx
Examples
Solve the following differential equations
(1)
1+y
dy
=
dx
2+x
Solution
This equation can be written in the separable form as
1 dy
1
=
1 + y dx
2+x
Integrating both sides ,we have
Z
Z
dy
dx
=
1+y
2+x
12
Hence,we have
ln(1 + y) = ln(2 + x) + C = ln(2 + x) + lnA
If we simplify ,we have
ln(1 + y) = lnA(2 + x)
Then we get
y(x) = A(2 + x) − 1
where A is an arbitrary constant
(2)
dy
= 2xy 2
dx
Solution
This equation can be written in the separable form as
y −2 dy = 2xdx
Integrating both sides,we have
Z
−2
y dy =
Z
2xdx
Hence,we have
−1
= x2 + C
y
If we simplify ,we get
y(x) = −
1
x2 + C
where C is an arbitrary constant of integration
13
3.2
Homogeneous Type
A differential equation of the form
dy
dx
= f (x, y) is said to be homogeneous of
degree n in x and y if and only if the function f(x,y) defined for all (x, y) ∈ R2
has the behaviour defined as follow;
f (tx, ty) = tn f (x, y) ∀t
Examples
(1) The differential equation
y
dy
x4
= 2y 3 e x −
dx
x + 3y
is a homogeneous equation of order 3 since the function
y
f (x, y) = 2y 3 e x −
x4
x + 3y
is such that
ty
f (tx, ty) = 2t3 y 3 e tx −
y
= t3 [2y 3 e x −
t4 x4
tx + 3ty
x4
]
x + 3y
= t3 f (x, y)
(2) The differential equation
dy
= x3 y 2 − 3x5
dx
is a homogeneous equation of order 5 since the function
f (x, y) = x3 y 2 − 3x5
14
is such that
f (tx, ty) = t3 x3 t2 y 2 − 3t5 x5
= t5 [x3 y 2 − 3x5 ]
= t5 f (x, y)
The solution of the homogeneous type required a change of variable y = vx
that reduces the equation to the variable separable type of equation.Once this
this is obtained,we can then use the separation of variable technique to solve
the problem.
Examples Solve the following differential equations
(1)
dy
x2 + y 2
=
dx
xy
Let
y = vx =⇒ v =
y
x
So that
dy
dv
=v+x
dx
dx
But
dy
x2 + v 2 x2
1 + v2
=
=
dx
x2 v
v
This means that
v+x
dv
1 + v2
=
dx
v
If we simplify this above equation ,we have
x
dv
1
=
dx
v
15
Integrating both sides of the equation ,we have
Z
vdv =
Z
dx
x
This gives
v2
= ln x + c =⇒ v 2 = ln x2 + 2c
2
After further simplification,we have
y 2 = x2 (ln x + 2c)
So that
√
y = x ln x2 + A
(2)
x + 3y
dy
=
dx
2x
Let
y = vx =⇒ v =
y
x
So that
dv
dy
=v+x
dx
dx
But
dy
x + 3xv
1 + 3v
=
=
dx
2x
2
This means that
v+x
dv
1 + 3v
=
dx
2
16
If we simplify the above equation,we have
x
1+v
dv
=
dx
2
Integrating both sides of the equation ,we have
Z
Z
2
dx
dv =
1+v
x
This gives
2 ln(1 + v) = ln x + c where c = ln A
After further simplification,we have
y
(1 + v)2 = Ax =⇒ (1 + )2 = Ax =⇒ (x + y)2 = Ax3
x
So that
√
y=
Ax3 − x
Exercise Solve the following differential equations:
dy
1. x2 + y 2 dx
= xy
dy
= xy − y 2
2. (x2 + xy) dx
dy
3. (x − y) dx
=x+y
3.3
The Exact Type
The equation we shall concern ourselves with here are either of the form:
y 0 = F (x, y)
(3.1)
M (x, y)dx + N (x, y)dy = 0
(3.2)
or
17
Example
(sin(x + y))dx + (x + 2y)dy = 0
may be written as
y0 =
dy
sin(x + y)
=−
dx
x + 3y
.
Definition 3.3.1. Let F be a function of 2 variables such that F has 1st partial
derivatives in the domain D. Then, the total differential equations of F ;
dF (x, y) =
∂F (x, y)
∂F (x, y)
dx +
dy ∀x, y ∈ D
∂x
∂y
(3.3)
Example:
Let F (x, y) = xy 2 + 2x3 y
so that
∂F (x, y)
dx = y 2 + 6x2 y,
∂x
∂F (x, y)
dy = 2xy + 2x3
∂y
Hence,
dF (x, y) = y 2 + 6x2 y + 2xy + 2x3
From (3.3),
dF (x, y) = M (x, y)dx + N (x, y)dy
then we can write
∂F
= M (x, y),
∂x
∂F
= N (x, y)
∂y
(3.4)
Theorem 3.3.1. Consider equation (3.2), where M and N have 1st partial
derivatives in (x, y) ∈ D. Then, the DE (3.2) is exact in D iff
∂M
∂N
=
∂y
∂x
∀ (x, y) ∈ D
(3.5)
18
But M =
∂F
∂x
and N =
∂F
,
∂y
then
∂
∂M
=
∂y
∂y
∂F
∂x
!
∂N
∂
=
∂x
∂x
∂F
∂y
!
=
∂ 2F
∂y∂x
=
∂ 2F
∂x∂y
Equation (3.6) confirms (3.5)
Consider
y 2 dx + 2xydy = 0
where
M (x, y) = y 2 ,
N (x, y) = 2xy
so that
∂M
= 2y,
∂y
∂N
= 2y
∂x
Since
∂M
∂N
=
= 2y
∂y
∂x
Then the given DE is an exact DE ∀ (x, y) ∈ D.
Exercise:
Verify whether the following differential equations are exact or not:
1. x2 dx + 2xydy = 0
2. ydx + 2xdy = 0
19
Theorem 3.3.2. Suppose (2.2) satisfies the differentiability requirement of
theorem (2.1) and its exact in (x, y) ∈ D, then the solution of the differential
equation (2.2) is given as
F (x, y) = c
(3.6)
where F is a function satisfying (2.4) and c is any arbitrary constant or function.
Example 1: Solve the DE
(3x2 + 4xy)dx + (2x2 + 2y)dy = 0
Solution:
M (x, y) = 3x2 + 4xy,
N (x, y) = 2x2 + 2y
and
∂M
= 4x,
∂y
∂N
= 4x
∂x
Hence, the DE is exact.
∂F
= 3x2 + 4xy
∂x
=⇒ F =
Z
3x2 + 4xydx
So,
F = x3 + 2x2 y + Φ(y),
(Φ is a f unction of
In other to know the value of Φ , we find
∂F
= 2x2 + Φ0 (y)
∂y
y)
20
But
∂F
= N (x, y) = 2x2 + 2y
∂y
Therefore
2x2 + 2y = 2x2 + Φ0 (y) =
∂F
∂y
and
Φ0 (y) = 2y,
=⇒ Φ(y) = y 2 + c
F = x3 + 2x2 y + y 2 + c
c at this level is optional.
Example 2: Solve the DE
(2x cos y + 3x2 y)dx + (x3 − x2 sin y)dy
Solution:
M (x, y) = 2x cos y + 3x2 y,
N (x, y) = x3 − x2 sin y
and
∂M
= −2x sin y + 3x2 ,
∂y
∂N
= 3x2 − 2x sin y
∂x
Hence the differential equation is exact
∂F
= 2x cos y + 3x2 y,
∂x
=⇒ F =
Z
2x cos y + 3x2 ydx
So
F = x2 cos y + +x3 y + φ(y)
21
In other to evaluate the value of Φ ,we find
∂F
= −x2 sin y + x3 + φ0 (y)
∂y
But
∂F
= N (x, y) = x3 − x2 sin y
∂y
Therefore
x3 − x2 sin y = −x2 sin y + x3 + φ0 (y)
and
φ0 (y) = 0,
=⇒ φ(y) = 0
Hence
F = x2 cos y + x3 y
Example 3: Solve the D.E
3x(xy − 2)dx + (x3 + 2y)dy = 0
Solution
M (x, y) = 3x2 y − 6x,
∂M
= 3x2 ,
∂y
N (x, y) = x3 + 2y
∂N
= 3x2
∂x
Hence the differential equation is exact
∂F
= 3x2 y − 6x,
∂x
=⇒ F =
Z
3x2 y − 6xdx
So
F = x3 y − 3x2 + φ(y)
22
In other to evaluate the value of Φ ,we find
∂F
= x3 + φ0 (y)
∂y
But
∂F
= N (x, y) = x3 + 2y
∂y
Therefore
x3 + 2y = x3 + φ0 (y)
and
φ0 (y) = 2y,
=⇒ φ(y) = y 2
Hence
F (x, y) = x3 y − 3x2 + y 2
The general solution F (x, y) = C ,
⇒ x3 y − 3x2 + y 2 = C
Exercise
(1) Determine which of the following differential equations are exact and
Solve those that are exact
(a.) (3x2 y + 2)dx − (x3 − y)dy = 0
(b.) (θ2 + 1) cos rdr + 2θ sin rdθ = 0
(c.) (2y sin x cos x + y 2 sin x)dx + (sin2 x − 2y cos x)dy = 0
(d.) ydx + (x2 y − x)dy = 0
(e.)
y
dx
x2
+ (y − x1 )dy
(2.) Determine the value of constant A which the equations below are exact
23
(a.) (x2 + 3xy)dx + (Ax2 + 4y)dy = 0
(b.) ( x12 +
3.4
1
)dx
y2
+ ( Ax+1
)dy = 0
y3
Non-Exact Equation To Exact Equation
Observe that the D.E
ydx + (x2 y − x)dy = 0
is not exact i.e
∂N
∂M
6=
∂y
∂x
However, if we multiply the equation by 1/x2 for instance, then we have
y
1
dx + (y − )dy = 0
2
x
x
which is exact (observe)
Generally, if
M (x, y)dx + N (x, y)dy = 0
is not exact, it is possible to make it exact by multiplying the equation with
a specific function that will make the equation exact.This function is called
Integrating factor.
Consider a Linear Differential equation of order 1 of the form
dy
+ Py = Q
dx
(3.7)
where P and Q are functions of x or constant
Equation(3.7) is not exact, but it can be solved using Integrating factor which
is always
e
R
pdx
24
Example 1
Solve the D.E
dy
+ 3y = e2x
dx
Solution
Compare this with (3.7), then, we observe that P = 3, Q = e2x then
I.F = e
R
3dx
= e3x
Multiply the given D.E by I.F, we have
e3x
dy
+ e3x 3y = e5x
dx
Observe that the LHS is the differential coefficient of ye3x i.e
d
dy
(ye3x ) = e3x
+ e3x 3y
dx
dx
Hence
d
(ye3x ) = e5x
dx
Integrating both sides, we have
3x
ye
=
Z
e5x dx =
e5x
+C
5
Divide both sides by e3x ,we have
y=
e2x
+ Ce−3x
5
25
Example 2
Solve the D.E
y0 − y = x
Solution
Compare with (3.7), P = −1, Q = x, then
R
I.F = e
−1dx
= e−x
Multiply the given D.E by I.F ,we have
e−x
dy
− e−x y = xe−x
dx
Observe that the LHS is the differential coefficient of ye−x i.e
d
dy
(ye−x ) = e−x
− e−x y
dx
dx
Hence
d
(ye−x ) = xe−x
dx
Integrating both sides,we have
ye−x =
Z
xe−x dx = −e−x (x + 1) + C
Using Integration by part
Divide both sides by e−x ,we have
y = −(x + 1) + Cex = Cex − x − 1
Example3 Solve
dy
+ ycotx = cosx
dx
26
Solution
Comparing with (3.7), p = cotx , Q = cosx,
R
cotxdx
R
cosx
dx
sinx
I.F = e
I.F = e
= eln sinx = sinx
Multiply the given D.E by I.F ,we have
sinx
dy
+ ycotxsinx = sinx
dy
2
ysinx= cosxsinxdx= sin2 x + c
R
y= sinx
+ ccosecx
2
Exercise
Solve the following D.E
dy
+ y = (x + 1)2
(1) (x + 1) dx
dy
(2) (1 − x2 ) dx
− xy = 1
dy
(3) x dx
− 5y = x7
27
3.5
Bernoulli Equation
This is a non linear special type of D.E which can be reduced to a linear
equation via logical transformation.
The equation of the form
dy
+ p(x)y = q(x)y n
dx
(3.8)
is called Bernoulli equation where n ∈ N
If n = 0 or n = 1, then(3.8) reduced to a linear equation which can be solved
using our previous method(s). It is therefore assume that n 6= 0, 1 for (3.8)
to retain its identity as Bernoulli D.E.The solution of this type of D.E require
the following transformation,
dy
+ p(x)y = q(x)y n
dx
Divide (3.8) by y n i.e
y −n
dy
+ p(x)y 1−n = q(x)
dx
(3.9)
Let
Z = y 1−n
(3.10)
dz
dy
= (1 − n)y −n
dx
dx
(3.11)
Then
Substitute (3.10) and (3.11) in (3.9), we have
1 dz
+ p(x)z = q(x)
1 − n dx
28
Multiple through by 1 − n,we have
dz
+ (1 − n)p(x)z = (1 − n)q(x)
dx
This can be written in the form
dz
+ Φ(x)z = Q(x)
dx
(3.12)
where
(1 − n)p(x) = Φ(x) and (1 − n)q(x) = Q(x)
. Observe that (3.12) is of the form (3.7) which is a linear D.E in variable x
and z
Example 1
Solve the D.E
dy y
+ = xy 2
dx x
Solution:
Divide through by y 2 ,
y −2
dy 1 1−2
+ y
=x
dx x
Let
z = y 1−2
..............(3.13)
Then
dz
dy
= −y −2
dx
dx
..........(3.14)
Substitute (3.13) and (3.14) into the given differential equation,we have
dz
1
− z = −x
dx x
29
The given equation has reduce to linear differential equation that be solve
using integrating factor i.e
I.F = e
R
−1
dx
x
−1
= elnx
= x−1 =
1
x
Multiply both sides of the last equation by this integrating factor,we have
1
1 dz
+ 2 z = −1
x dx x
Note that the left hand side of the above equation satisfy the differential coefficient
z
d
(+ ) = −1
dx x
Integrating both sides,we have
Z
z
= − dx = −x + c
x
Then
z = −x2 + cx
Therefore
y = (−x2 + cx)− 1
Example 2
Solve the D.E
x2 y − x3
dy
= y 4 cos x
dx
Solution:
This equation can be re-arrange as
dy
1
−y 4 cosx
− y=
dx x
x3
30
Divide through by y 4 , we have
y −4
dy
1
cosx
− y −3 = 3 ....... +
dx x
x
Let
z = y 1−4 = y −3 .......... ∗
So that the derivative
dz
dy
= −3y −4 ..........
dx
dx
∗∗
Substitute ∗ and ∗∗ in equation +, and divide both sides of the equation by
−3 we have
3
dz
+ z = −3x−3 cosx
dx x
This is a linear differential equation with variable z and x
Where
I.F = e
R
3
dx
x
= x3
Multiple the linear equation by the integrating factor ,we have
x3
dz
+ 3x2 z = 3cosx
dx
Observe that the RHS of above equation is the derivative of
So that
d 3
(x z) = 3 cos x
dx
Integrate both sides, we have
x3 z =
Z
3cosxdx
d
(x3 z)
dx
31
z=
3 sin x + c
x3
and
x3
x3
y3 = R
=
[ 3cosxdx] + c
3sinx + c
Therefore,
y=(
1
x3
)3
3sinx + c
Exercise
Solve the D.E
2y −3
dy
= y 4 e3x
dx
Chapter 4
Techniques for Solving Second
order Linear and Non Linear
O.D.E
4.1
Homogeneous Non Linear With Missing
Dependent/Independent Variable
An equation of the form
0
00
F (x, y, y , y ) = 0
(4.1)
is called a second oder O.D.E involving 2nd order derivative of the unknown
function y(x) as the highest order. We consider this type of equation under 2
cases
Case 1
The case when dependent variable 0 y 0 is missing is of the form
0
00
G(x, y , y ) = 0
Suppose y is the solution of (4.2), then we can set v =
32
(4.2)
dy
dx
such that V becomes
33
a solution of the first order equation
0
G(x, v, v ) = 0
(4.3)
If (4.3) is solvable, then the solution of (4.2) can be obtained by integrating
v=
dy
dx
Example: Solve the diferential equation
x
dy
dy
d2 y
= 2([ ]2 − )
2
dx
dx
dx
This is of the form (4.2) with missing variable y i.e the dependent variable
Solution:
Let v =
dy
dx
then the equation becomes
x
dv
= 2(v 2 − v)
dx
Seperating variable,we have
dx
dv
=
2
v2 − v
x
Appling partial fraction,we have,
(
1
1
dx
− )dv = 2
v−1 v
x
Integrating both sides,we have
ln(v − 1) − lnv = 2lnx + lnc
Simplify,we have
v−1
= cx2
v
34
But
v=
dy
dx
then
1−
−1
= cx2
v
=⇒
dy
1
=
dx
1 − c1 x 2
By integration
y=
1 + ax
1
ln(
) + c2
2a 1 − ax
if C1 ≥ 0
And
1
y = tan−1 bx + c if C1 ≤ 0
b
Case 2
The case when independent variable is missing i.e
G(y, y 0 , y 00 ) = 0
(4.4)
Let
v=
d2 y
dv
dv dy
dv
=
= .
=v
2
dx
dx
dy dx
dy
dy
,
dx
Then we ca write 4.4 as
G(y, v, v
dy
)=0
dx
(4.5)
(4.5) is a first order D.E for dependent variable v. If (4.5) is solvable then,
(3.7) has a solution by solving
dy
= v(y)
dx
Example: Solve the D.E
d2 y
dy
dy
= ( )2 + 2
2
dx
dx
dx
(4.6)
35
Solution:
Since x is missing , then v =
dy
,
dx
and
yv
d2 y
dx2
dv
= v dy
, so (4.6) becomes
dv
= v 2 + 2v
dy
This is can be reduce to
y
dv
=v+2
dy
Seperate the variable,we have
dy
dv
=
v+2
y
Solving, we have
lnv + 2 = lny + lnc1 = lnc1 y
v + 2 = cy
and v = c1 y − 2
i.e
dy
= c1 y − 2
dx
Seperate the variables ,we have
dy
= dx
c1 y − 2
Integrate both sides,we have
1
ln(c1 y − 2) = x + c2
c1
so that
C1 y − 2 = ec1 x+c = Aec1 x
(4.7)
36
Where C = c1 c2 .So
C1 y = Aec1 x + 2
Therefore,
y=
A c1 x
2
e +
c1
c1
Exercise
Solve the differential equations
2
dy 2
d y
(1) y dx
2 + ( dx ) = 0
2
d y
dy 2
(2) 2x dx
2 = ( dx ) − 1
4.2
Second order Homogeneous Linear Differential Equation
Generally,second order linear Ordinary Differential Equation is of the form
a0 (x)
d2 y
dy
+ a1 (x) + a2 (x)y = f (x)
2
dx
dx
(4.8)
where a0 , a1 ,a2 are either constant or functions of x and f is a functions of (x)
If a0 , a1 and a2 are constant and f = 0, then (4.8) becomes Homogeneous
Second Order Linear Differential Equation with constant coefficients i.e
a0
d2 y
dy
+ a1
+ a2 y = 0
2
dx
dx
If y1 (x) and y2 (x) are the solutions of (4.9),so also is
y(x) = y1 (x) + y2 (x) a solution
(4.9)
37
Procedure for solution of (4.9)
If a0 = 0 in (4.9),we have the 1st order ordinary differential equation provided
a0 and a0 are not equal to zero,so
a1
dy
+ a2 y = 0
dx
Divide through by a1 ,we have
dy a2
+ y=0
dx a1
Let
a2
a1
= k ,then we have
dy
= −ky
dx
Seperate the variable and intergrate both sides,we have
lny = −kx + c,
y = e−kx+c = Ae−kx
If − k = m then y = Aemx
If y is a solution of (4.9),then
dy
= Amemx ,
dx
d2 y
= Am2 emx
dx2
Substitute these into equation(4.9),we have
a0 Am2 emx + a1 Amemx + a2 Aemx = 0
so
Aemx (a0 m2 + a1 m + a2 ) = 0
Hence
a0 m2 + a1 m + a2 = 0
(4.10)
38
Equation (4.10) is called the Auxilliary or Characteristics Equation of the
differential equation (4.9)
The solution of (4.10) can be classify into three i.e
(a) Real and distinct solutions (roots)
(b) Real and equal solution (roots)
(c) Complex solution (roots)
Real and Distinct roots
Example 1
Consider the differential equation
dy
d2 y
−
3
+ 2y = 0
dx2
dx
The auxilliary equation is
m2 − 3m + 2 = 0
Solve this,we have
(m − 1)(m − 2),
m1 = 1,
m2 = 2
Since the roots of the auxilliary equation are distinct,then the general solution
is
y(x) = A1 em1 x + A2 em2 x = A1 ex + A2 e2x
Example 2
Given the differential equation
d2 y
dy
= −4 + 5 = 0
2
dx
dx
39
with auxilliary equation
m2 + 4m − 5 = 0
solve this ,we have
m1 = 1,
m2 = 5
So the general solution is given by
y(x) = A1 ex + A2 e−5x
Real and Equal (Repeated) roots
Example
Consider tge differential equation
d2 y
dy
−
6
+ 9y = 0
dx2
dx
The auxilliary equation is given as,
m2 − 6m + 9 = 0
solve this ,we have
m1 = 3,
m2 = 3
so the general solution is expected to be
y(x) = A1 e3x + A2 e3x
= (A1 + A2 )e3x
= Ae3x
(Since the addition of two constant is another constant)
But we know that every second order differential equation must have two arbotrary constants,so there must be anotherterm containing a second constant.
40
Thus,we must have two linearly independent solutions as
y(x) = Ae3x + Bxe3x
(A + Bx)e3x
Generally,the general solution of a differential equation whose auxilliary roots
are real and repeated is
y(x) = (A + Bx)e3x
Complex Conjugate roots
If the roots of the auxilliary equation are complex conjugate roots such that
m1 = α + iβ,
m2 = α − iβ
then the solution of (4.9) are
y1 (x) = A1 e(α+iβ)x ,
y2 (x) = A2 e(α−iβ)x
so that we have
y(x) = A1 e(α+iβ)x + Ae(α−iβ)x
= A1 eαx eiβ)x + A2 eαx e−iβ)x
eαx (A1 eiβ)x + A2 e−iβ)x )
eαx (A1 (cos β + i sin βx) + A2 (cos β + i sin βx)
since
eiθ = cos θ + i sin θ,
e−iθ = cos θ − i sin θ
therefore
y(x) = eαx [A1 cos βx + A2 cos βx + iA1 sin βx − iA2 sin βx]
41
= eαx [A cos βx + B sin βx]
where A1 +A2 = A and i(A1 −A2 ) = B where A and B are arbitrary constants
Example
Find the general solution of the differential equation
y 00 − 2y 0 + 10y = 0
Solution
The auxilliary equation associated with above differential equation is given by
m2 − 2m + 10 = 0
Solve this ,we have the auxilliary roots as
m1 = 1 + 3i,
m2 = 1 − 3i
This means that α = 1 and β = 3
Therefore ,the general solution of the differential equation is given by
y(x) = ex [A cos 3x + B sin 3x]
Exercise
Solve the following differential equations
(1) y 00 − 5y 0 + 6 = 0
(2) y 00 − 8y 0 + 16y = 0
(3) y 00 + cy = 0 if
c>0
Initial Value Problem (IVP)
42
The initial value problem is a difeferntial equation with initial condition(s).This
initial condition(s) helps us to find the particular solution of the differential
equation after obtaining the general solution.We shall illustrate this with the
example
Example
Solve the Initial Value Problem
d2 y
dy
− 6 + 25y = 0,
2
dx
dx
y(0) = −3,
y 0 (0) = −1
Solution
The auxilliary equation of the above D.E is
m2 − 6m + 25 = 0
Solve this ,we have
m1 = 3 + 4i,
m1 = 3 − 4i
i.e α = 3 ,and β = 4
The general solution therefore is
y(x) = e3x (A1 sin 4x + A2 cos 4x)
We can find A1 and A2 by using the initial condition
−3 = e0 (A1 sin 0 + A2 cos 0)
−3 = A2
y 0 (x) = e3x [(3A1 − 4A2 ) sin 4x + (3A2 + 4A1 ) cos 4x)
−1 = e0 [(3A1 − 4A2 ) sin 0 + (3A2 + 4A1 ) cos 0)
43
−1 = (3A2 + 4A1 )
−1 = (4A1 − 9)
A1 = 2
Therefore ,the particular solution of the differential equation is
y(x) = e3x (2 sin 4x − 3 cos 4x)
4.3
Second Order Non Homogeneous Differntial Equation
Consider the differential equation of the form
a0 (x)
d2 y
dy
+ a1 (x) + a2 (x)y = f (x)
2
dx
dx
(4.11)
If f (x) is zero,the equation is homogeneous as previously stated,but if f (x)
is not zero,then the equation is Non-Homogenous.The solution of this non
homogeneous equation is the addition of the solution of the homogeneous part
and the non homogeneous part i.e
y(x) = yc (x) + yp (x)
where yc (x) is called the complimentary equation and yp (x) is called the particular integral
We shall solve this type of differential equation with the following techniques
namely
• Method of Undertermined Coefficient
• Method of Variation of Parameter or Constant
44
• The Operator D method
We shall examine these methods one after the other as follows
Method of Undertermined Coefficient
We shall consider when the function f (x) is of the form
(a) Pn (x)
(b) Pn (x)eax
(c) Pn (x) sin βx
(d) Pn (x) cos βx
(e) Pn (x) sinh x
(f) Pn (x) cosh x
We summarize these and the corresponding particular integral on the table
below
f (x)
Pn (x)
Pn (x)eax
Pn (x) sin βx
Pn (x) cos βx
yp (x)
a0 + a1 x + a2 x2 ....an xn
(a0 + a1 x + a2 x2 ....an xn )eax
(a0 + a1 x + a2 x2 ....an xn )eax sin bx + (c0 + c1 x + c2 x2 ....cn xn )eax cos bx
(a0 + a1 x + a2 x2 ....an xn )eax sin bx + (c0 + c1 x + c2 x2 ....cn xn )eax cos bx
This require that we solve the homogeneous part first to get the complimentary function and then the part with f (x) 6= 0 to get the paticular integral
Example Solve the differential equation
y 00 − 5y 0 + 6y = x2
45
This has auxilliary equation
m2 − 5m+ = 0
Solve this to get
m = 2,
m=3
Therefore the complimentary function is
y(x) = A1 e2x + A2 e3x
To find the particular integral,we assume the general form of the RHS of the
given problemwhich is a polynomial of order 2. so the that,
yp (x) = Cx2 + Dx + E
where C,D and D are unknown constant to be determine
Next,we find the first and the second derivative of the particular intgral and
the substitute them into the given problem to find the unknown constants
so
yp0 = 2Cx + D,
yp00 = 2C
Substitute these into the given D.E, we have
2C − 5(2Cx + D) + 6(Cx2 + Dx + E) = X 2
6Cx2 + (6D − 10C)x + (2C − 5D + 6E) = x2
46
Equating corresponding coefficients,we have
6C = 1,
=⇒
6D − 10C = 0,
2C − 5D + 6E = 0,
C=
=⇒
1
6
5
18
19
E=
108
D=
=⇒
So the particular integral for the given D.E is given by
5
19
1
yp (x) = x2 + x +
6
18
108
And the general soution is
1
5
19
y(x) = A1 e2x + A2 e3x + x2 + x +
6
18
108
where A1 and A2 are arbitrary constant that can be determined if the initial
conditions are given.
Example
Solve the D.E
y 00 + y 0 = xe2x
Solution
The auxilliary equation of the homogeneous part is
m2 + 1 = 0
If we solve,we get
m1 = i,
m2 = −i
So the complimentary function is
yc (x) = A1 sin x + A2 cos x
47
But f (x) is of the form Pn (x)eax where Pn (x) is a polynomial of degree one,so
the particular integral is
yp (x) = e2x (a + bx)
Differentiate the P.I twice and substitute into the given D.E,we have
y 0 = e2x (2a + 2bx + b)
and
y 00 = e2x (4a + 4bx + 4b)
substitute these we have,
e2x (4a + 4bx + 4b) + e2x (a + bx) = xe2x
simplify we have,
e2x (4a + 4bx + 4b + a + bx) = xe2x
so
(5a + 4b + +5bx) = x
Equating corresponding coefficients ,we have
5a + 4b = 0,
5b = 1
Solve these simultaneously ,we have
a=
−4
,
25
b=
1
5
This implies that the particular integral is
yp (x) = e2x (
−4 1
+ x)
25
5
48
Therefore the general solution of the D.E is given as
A1 sin x + A2 cos x + e2x (
−4 1
+ x)
25
5
Method of Variation of Parmeters
If f (x) is not in any of the form described under the method of undetermined
coefficient,then we require another method which is method of variation of
parameter,which is more general Consider
y 00 + ay 0 + by = f (x)
(4.12)
If we have two linearly independent solutions y1 and y2 of the homogeneous
equation
y 00 + ay 0 + by = 0
so that a general solution is given by
yc (x) = C1 (x)y1 (x) + C2 y2
with C1 and C2 satisfying the differential
C10 (x) = −
C20 (x) = −
f (x)y2 (x)
)
− y2 (x)y10 (x
y1 (x)y20 (x)
f (x)y1 (x)
)
− y2 (x)y10 (x
y1 (x)y20 (x)
provided the denominator is not zero, where
y1 (x)y20 (x) − y2 (x)y10 (x) = |W (y1 , y2 )|
49
is the Wronskian
Therefore the general solution will be
y(x) = yc (x) + yp (x)
Example
Solve the differential equation
y 00 + y = tan x
Solution
If we follow the previous discussion,we see that the general solution of this
equation is given as
yc (x) = A1 cos x + A2 sin x
where
y1 (x) = cos x and y2 (x) = sin x
so
C10 (x) = −
tan x sin x
= cos x − sec x
cos x cos x + sin x sin x
Integrate both sides we have
C1 =
Z
cos x − sec xdx = sin x − ln| sec x + tan x|
Also,
C20 (x) = −
tan x cos x
= − sin x
cos x cos x + sin x sin x
Integrate both sides we have
C2 =
Z
− sin xdx = cos x
50
This implies that
yp (x) = (sin x − ln| sec x + tan x|) cos x + cos x sin x
Therefore the general solution of the D.E is
y(x) = A1 cos x + A2 sin x + (sin x − ln| sec x + tan x|) cos x + cos x sin x
4.4
Operator D Method
Chapter 5
Higher Order Homogeneous and
Non Homogeneous Differential
Equation
5.1
Homogeneous Type
The process of solving this type of differential equation is similar to other
homogeneous equation
Example 1
Solve the D.E
y 000 + 14y 00 + y 0 − 6y = 0
Solution
The auxilliary equation of this D.E is
m3 + 14m2 + m − 6 = 0
If we solve this ,we have
m1 = 1,
m2 = −2,
51
m3 = −3
52
Therefore the general solution is
y(x) = c1 ex + c2 e−2x + c3 e−3x
Example 2
Solve the D.E
y iv + 16y 000 + 96y 00 + 256y 0 + 256y = 0
Solution
The auxilliary equation of this D.E is
m4 + 16m3 + 96m2 + 256m + 256 = 0
If we solve this ,we have
m1 = −4,
m2 = −4,
m3 = −4,
m4 = −4
Therefore the general solution is
y(x) = c1 e−4x + c2 xe−4x + c3 x2 e−4x + c4 x3 e−4x
5.2
Non Homogeneous Type
Consider the nth order non homogeneous equation
y n + a1 (x)y n−1 + a2 (x)y n−2 + .... + an (x)y = f (x)
(5.1)
and it associated homogeneous equation
y n + a1 (x)y n−1 + a2 (x)y n−2 + .... + an (x)y = 0
(5.2)
53
If y1 , y2 ....yn are linearly independent solutions of (5.2),then the solutiony(x)
of (5.2) is expressed as
y(x) = c1 y1 (x) + c2 y2 (x) + .... + cn yn (x)
where c1 , c2 ,.....cn are constant
If y1 , y2 ....yn are independent solutions of (5.2),then the Wronskian
W (y1 , y2 ....yn ) = |y1 , y2 ....yn y10 , y20 ....yn0 ....y1n−1 , y2n−1 ....ynn−1 | |= 0
(5.3)
Using the method of Variation of Parameter,we find a solution (Partivular
Integral) of the form
yp (x) = c1 (x)y1 (x) + c2 (x)y2 (x) + .... + cn (x)yn (x)
Differentiating ,we have
yp0 (x) = c1 (x)y10 (x)+c01 (x)y1 (x)+c2 (x)y20 (x)+c02 (x)y2 (x)+....+cn (x)yn0 (x)+c0n (x)yn (x)
= (c1 (x)y10 (x)+c2 (x)y20 (x)+...+cn (x)yn0 (x))+(c01 (x)y1 (x)+c02 (x)y2 (x)+....+c0n (x)yn (x))
If we set
(c01 (x)y1 (x) + c02 (x)y2 (x) + .... + c0n (x)yn (x)) = 0
Then
yp0 (x) = c1 (x)y10 (x) + c2 (x)y20 (x) + ... + cn (x)yn0 (x)
Differentiating one more time ,we have
yp00 (x) = c1 (x)y100 (x)+c01 (x)y10 (x)+c2 (x)y200 (x)+c02 (x)y20 (x)+....+cn (x)yn00 (x)+c0n (x)yn0 (x)
= (c1 (x)y100 (x)+c2 (x)y200 (x)+....+cn (x)yn00 (x))+(c01 (x)y10 (x)+c02 (x)y20 (x)+....+c0n (x)yn0 (x))
54
If we set
(c01 (x)y10 (x) + c02 (x)y20 (x) + .... + c0n (x)yn0 (x)) = 0
then we have
yp00 (x) = c1 (x)y100 (x) + c2 (x)y200 (x) + .... + cn (x)yn00 (x)
If we continue this way,we set
c01 (x)y1k (x) + c02 (x)y2k (x) + .... + c0n (x)ynk (x) = 0,
k = 0, 1, 2...(n − 1)
then we have
ypk (x) = c1 (x)y1k (x) + c2 (x)y2k (x) + .... + cn (x)ynk (x),
k = 0, 1, 2...(n − 1)
and
ypn−1 (x) = c1 (x)y1n−1 (x) + c2 (x)y2n−1 (x) + .... + cn (x)ynn−1 (x)
so that
ypn (x) = (c1 (x)y1n (x)+c2 (x)y2n (x)+....+cn (x)ynn (x))+(c01 (x)y1n−1 (x)+c02 (x)y2n−1 (x)+....+c0n (x)ynn−1 (x))
This represent the nth derivative of yp (x)
With all these,we have suceeded in forming a system of n equations in 4n4
unknowns i.e
c01 (x)y1 (x) + c02 (x)y2 (x) + .... + c0n (x)yn (x) = 0
c01 (x)y10 (x) + c02 (x)y20 (x) + .... + c0n (x)yn0 (x) = 0
.
.
.
c01 (x)y1n−1 (x) + c02 (x)y2n−1 (x) + .... + c0n (x)ynn−1 (x) = 0
55
The determinant of this system is called WRONSKIAN which is non zero since
the funcyions y1 , y2 , ....yn are linearly independent
Example
Solve the differential equation
d3 y
dy
− 2y = 3e−x
+
3
dx3
dx
Solution
The homogeneuos part has an auxilliary equation given as
m3 + 3m − 2 = 0
Solve this,we have the roots
m1 = −1,
m2 = −1,
m3 = 2
and this will have a complimentary function given as
yc (x) = A1 e−x + A2 xe−x + A3 e2x
where
y1 = e−x ,
y2 = xe−x ,
so the wronskian is given by
w(x)=
e−x
xe−x
e2x
−e−x (1 − x)e−x 2e2x
e−x (x − 2)e−x 4e2x
w(0)=
1 0 1
−1 1 2
1 −2 1
=9
=0
y3 = e2x
56
By Crammer’s rule,
0
xex
e2x
0 (1 − x)e−x 2e2x
−x
3e
(x − 2)e−x 4e2x
w1 (x)=
e−x 0 e2x
−e−x 0 2e2x
e−x e−x 4e2x
e−x
xe−x
−x
−e
(1 − x)e−x
e−x (x − 2)e−x
w2 (x)=
w3 (x)=
C10 =
w1
w
=
9x−3
9
C20 =
w2
w
=
−9
9
C30 =
w3
w
=
3e−3x
9
C1 =
x2
2
−
x
3
=
9x
9
−
3
9
= 9x − 3
= −9
0
0
3ex
=x−
= 3ex
1
3
=1
,
=
e−3x
3
C2 = −x , C3 =
−1 −3x
e
9
So,
y(x) = C1 (x)y1 (x) + C2 (x)y1 (2) + C3 (x)y1 (3)
=
x2
2
2
−
x
3
= x2 e−x −
e−x -x(xe−x ) -
Therefore,
x
3
+
1
9
e−x
1 −3x
e
9
e2x
57
y(x) = yc + yp
A1 e−x + A2 xe−x + A3 e2x −
x2 −x
e
2
−
x
3
+
1
9
e−x
Chapter 6
Numerical Solution
6.1
Error and Interpolation
Error/Types of Error
(1.) Gross Error or Blunder: This is the error committed whwn a wrong
answer is written for a correct one. e.g. Writing 0.5971 instead of 0.5791.
(2.) Truncation Error This is the error committed when an infinite process
is replaced by a finite process.e.g. The expansion of
1
(1 + x) 2 = 1 +
x x2
−
+ .......
2
8
(3.) Round-Off Errors: This is the error committed when certain arithmetic calculation obtain as fraction is required to be expressed as decimal, leading to approximation e.g. 1/3 = 0.333,. . . , 3. . .
Let E = Error,
X = true value , and X ∗ = approximated value.
then, Absolute Error = |X − X ∗ |
58
59
Relative Error =
Actual Error
E
X − X∗
X∗
=
=
=1−
T rue V alue
X
X
X
If X is a real number which in general has an infinite decimal representation,
we say that x has been rounded off to a d-decimal places written as x(d) i.e.
|E| = x − x(d) ≤
× 10−d
1
2
x(2) = 0.14
If x = 0.1428571,
|E| = x − x(2) = |0.1428571 − 0.14| = 0.0028571
0.0028571 ≤
1
2
× 10−d ≤
1
2
×
1
100
≤
1
200
≤ 0.005
x(3) = 0.333
If x = 0.333333
|E| = |0.333333 − 0.333| = 0.000333
0.000333 ≤
1
2
× 10−3 =
Hence, x − x(d) ≤
1
2
1
2
×
1
1000
=
1
2000
= 0.0005
× 10−d
Errors Propagation
Let X and Y be the real values and let X ∗ and Y ∗ be their respective approximation with εX and εY as their respective error. then,
X∗ = X − ε
and
Y∗ =Y −Y
εX = X − X ∗
and
εY = Y − Y ∗
(1.) Addition
X ∗ + Y ∗ = (X − εX ) + (Y − εY )
= X − εX + Y − εY
=(X + Y ) − (εX + εY ) (εX + εY ) = (X + Y ) − (X ∗ + Y ∗ )
The sum is the addition of the errors
60
(2.) Subtraction
X ∗ − Y ∗ = (XεX ) − (Y − εY )
=X − εX − Y + εY
=(X − Y ) − (εX − εY ) = (X − Y ) + (εY − εX )
⇒ (εX − εY ) = (X − Y ) − (X ∗ − Y ∗ )
The error is the difference of the error
(3.) Multiplication
X ∗ × Y ∗ = (X − εX )(Y − εY ) = XY − XY − Y εX + εX εY
Neglecting εX εY , then
X ∗ × Y ∗ = XY − Xεy − Y εX
XY − X ∗ Y ∗ = XεY + Y εX
But Relation Error =
Actual Error
XY − X ∗ Y ∗
XεY
Y εY
εY
εX
=
=
+
=
+
T rue V alue
XY
XY
XY
Y
X
Therefore, the relative error in the product equals the sum of the relative
error in the number we are multiplying.
(4.) Division
X∗
Y∗
=
X−εX
Y −εY
X
X∗
X
X − εX
X(Y − εY ) − Y (X − εX )
− ∗ =
−
=
Y
Y
Y
Y − εY
Y (Y − εY )
XY − XεY − XY + Y εX
Y εX − XεY
=
Y (Y − εY )
Y (Y − εY )
Relative Error
Y X −X Y
1
=
×
P
Y (Y − Y )
x/y
P
P
61
=
X
X
1
Y
×
(Y
−X
)
×
P
Y2−Y Y
X
x
Y
1
Y2
= 2
P ·
Y −Y Y
X
P !
P
XY
−
X
X
1
Y2
= 2
P ·
Y −Y Y
X
P
1
= 2
P ·
Y −Y Y
=
1
Y 2 (1 −
=
1
(1 −
·
P
Y
Y
P
X
−
X
P
)
Y
Y
−Y
P
X
X
)
!
X
Y
Y2
X
·Y2
P
X
Y
X
−
P Y
Y
P !
XY
−
X
Y
P Y
Y
P
= Plim
y
→0
Y
Y
P
=
X
X
→0
P
−
Y
Y
Therefore, the relative error in the division equals the difference of the relative
error of the respective number we are dividing.
Interpolation
62
Given n+1 data with (xi , fi ),
i = 0, 1, ..., n, we seek a polynomial Pn (x)
which takes fi at every xi i.e Pn (xi ) = fi
Pn (x0 ) = f0 ,
Pn (x1 ) = f1 ,
...,
Pn (xn ) = fn
The polynomial Pn (x) is called INTERPOLATING POLYNOMIAL and fi are
0
certain mathematical functions Pn (xi ) is an estimate of fi at respective xis
The process of obtaining the Pn (xi ), the estimate of fi (xi ) is called INTERPOLATION,If x of interest lies within xi and it iss called EXTRAPOLATION, if
x of interest does not lies within xi
Special Cases:
1. Linear interpolation: This is interpolation by means of straight line
through two points (x0 , f0 ) & (x1 , f1 ) given by
P1 (x) = f0 + (x − x0 )f (x0 , x1 )
where f (x0 , x1 ) =
f1 −f0
x1 −x0
Note: (1) P1 (x0 ) = f0
P1 (x1 ) = f0 + (x1 − x0 )
f1 − f0
x1 − x0
P1 (x1 ) = f0 + f1 − f0 = f1
Exercise:
(1) Estimate the population of Nigeria in 1999 given that
Year
1990 2006
Population(million) 120 150
63
Answer is 136.875 million
(2) Find ln 9.2 from ln 9.0=2.1972, ln 9.5=2.2513
2. Quadratic interpolation:
This is interpolation of second order polynomial with a curve through
(x0 , f0 ),
(x1 , f1 ),
(x2 , f2 )
given by
P2 (x) = f0 + (x − x0 )f (x0 , x1 ) + (x − x0 )(x − x1 )f (x0 , x1 , x2 )
where
f (x0 , x1 , x2 ) =
f (x1 , x2 ) − f (x0 , x1 )
,
x2 − x0
f (x1 , x2 ) =
f2 − f1
x2 − x1
Exercise:
(1) verify that P2 (X0 ) = f0 and P2 (X1 ) = f1
(2) Compute the quadratic interpolation for f (0.9) from f (0.5) = 0.479, f (1.0) =
0.841, f (2.0) = 0.909
(3) Compute the quadratic interpolation for Γ(1.01) and Γ(1.03) from
Γ(1.00) = 1.000, Γ(1.02) = 0.989, Γ(1.04) = 0.978
(4) Obtain the polynomial for ln8.0 = 2.0794, ln9.0 = 2.1972, ln9.5 =
2.2573
64
3. Lagrange Interpolation
This invplves the interpolation of polynomial with order n > 2
Given a table of n-values [(x0 , f0 ), ..., (xn , fn )] of a function f (xj ) evaluated at x = xj ,
j = 1, 2, ..., n, we require to evaluate the value of f (x)
when x is not tabulated. The process is to approximate f (x) by a function which passes through each of the tabulated point. This is carried
out by the Lagrange interpolation formula.
f (x) = Ln (x) =
n
X
lk (x)
k=0 lk (xk )
f (xk )
where
lk (x)
(x − x0 )(x − x1 )...(x − xj−1 )(x − xj+1 )...(x − xn )
= lj (x) =
,
lk (xk )
(xj − x0 )(xj − x1 )...(xj − xj−1 )...(xj − xn )
L3 (x) =
=
j = 0, 1, ..., n
l0 (x)
l1 (x)
l2 (x)
l3 (x)
f0 +
f1 +
f2 +
f3
l0 (x0 )
l1 (x1 )
l2 (x2 )
l3 (x3 )
(x − x1 )(x − x2 )(x − x3 )
(x − x0 )(x − x2 )(x − x3 )
(x − x0 )(x − x1 )(x
f (x0 )+
f (x1 )+
(x0 − x1 )(x0 − x2 )(x0 − x3 )
(x1 − x0 )(x1 − x2 )(x1 − x3 )
(x2 − x0 )(x2 − x1 )(x
Exercise:
(1) Find the lagrange interpolation polynomial for the data
x
0 2 3 5
f(x) 1 3 2 5
x0 = 0, x1 = 2, x2 = 3, x3 = 5
f0 = 1, f1 = 3, f2 = 2, f3 = 5
65
Answer:
3 3
x
10
−
15 2
x
6
+
62
x
15
+1
(2) Let f (x) = lnx. Estimate the value of ln(0.6) from the table
x
0,4
0.5
0.7
0.8
f(x) -0.9165 -0.6931 -0.3567 -0.2231
Answer = −0.5100
(3) Find ln9.2 using the table below
x
9.0
9.5
10.0
11.0
ln x 2.69722 2.25129 2.30259 2.39790
Answer = 2.21920