LECTURE 6 (Week 6)

Dr. Rehan Ali Shah

Differential Equation
(BSI-231) for Industrial
Engineering (Jalozai)
 Contents

 Cauchy-Euler Equations

 Undetermined Coefficients Method

 Nonlinear Equations

CH3_2
 3.6 Cauchy-Eulaer Equation
 Form of Cauchy-Euler Equation
n n 1
nd y n 1 d y dy
an x n
 an1x n1
   a1x  a0 y  g ( x)
dx dx dx

 Method of Solution
We try y = xm, since
k
kd y mk
ak x  a x k
m ( m  1)( m  2 )  ( m  k  1) x
dx k k

 ak m(m  1)( m  2)(m  k  1) x m

CH3_3
 An Auxiliary Equation

 For n = 2, y = xm, then

am(m – 1) + bm + c = 0, or
am2 + (b – a)m + c = 0 (1)

 Case 1: Distinct Real Roots

(2) y  c1 x m1  c2 x m2

CH2_4
 Example 1
2
2 d y dy
Solve x 2
 2x  4 y  0
dx dx
Solution:
We have a = 1, b = -2 , c = -4
m2 – 3m – 4 = 0, m = -1, 4,
y = c1x-1 + c2x4

CH3_5
 Case 2: Repeated Real Roots
m1
 Using (5) of Sec 3.2, we have y2  x ln x
Then
m1 m1
y  c1x  c2 x(3)ln x

CH3_6
 Example 2
2
2 d y dy
Solve 4 x 2
 8x  y  0
dx dx
Solution:
We have a = 4, b = 8, c = 1
4m2 + 4m + 1 = 0, m = -½ , -½
y  c1x 1/2  c2 x 1/2 ln x

CH3_7
 Case 3: Conjugate Complex Roots

 Higher-Order: multiplicity
x m1 , x m1 , x m1 (ln x) 2 ,  , x m1 (ln x) k 1
 Case 3: Conjugate Complex Roots
m1 =  + i, m2 =  – i,
y = C1x( + i) + C2x( - i)
Since
xi = (eln x)i = ei ln x = cos( ln x) + i sin( ln x)
x-i = cos ( ln x) – i sin ( ln x)
Then
y = c1x cos( ln x) + c2x sin( ln x)
= x [c1 cos( ln x) + c2 sin( ln x)] (4)
CH3_8
 Example 3
1
Solve 4 x y  17 y  0, y (1)  1, y ' (1)  
2
2
Solution:
We have a = 4, b = 0 , c = 17
4m2 − 4m + 17 = 0, m = ½ + 2i
1/ 2
y  x [c1 cos(2 ln x)  c2 sin(2 ln x)]
Apply y(1) = -1, y’(1) = 0, then c1 = -1, c2 = 0,
1/2
y   x cos(2 ln x)
See Fig 3.15.

CH3_9
Fig 3.15

CH3_10
 Example 4
3 2
3 d y 2 d y dy
Solve x 3
 5x 2
 7x  8y  0
dx dx dx
Solution:
Let y = xm,
2
dy d y
 mx , 2  m(m  1) x m2 ,
m1
dx dx
d3y m3
 m ( m  1 )( m  2) x
dx3
Then we have xm(m + 2)(m2 + 4) = 0
m = -2, m = 2i, m = -2i
y = c1x-2 + c2 cos(2 ln x) + c3 sin(2 ln x)
CH3_11
 Solution of non-homogeneous DE
  
 Solution of non-homogeneous DE
 
 The Method of Undetermined Coefficients
 
 
 

// /
ay  by  cy
p p p
 Caution!

 In addition to the form of the input function g (x ) , the educated guess for
must take into consideration the functions that make up the complementa
ry function
y
c
.
 No function in the assumed y p must be a solution of the associated homogen
eous differential equation. This means that the assumed y
should
p not con
tain terms that duplicate terms in . y
c
 Rule of Case 1:

 No function in the assumed yp is part of yc

Table 3.1 shows the trial particular solutions.
g (x) Form of yp
1. 1 (any constant) A
2. 5x  7 Ax  B
3. 3x 2  2 Ax 2  Bx  C
4. x3  x  1 Ax3  Bx 2  Cx  E
5. sin 4 x A cos 4 x  B sin 4 x
6. cos 4 x A cos 4 x  B sin 4 x
7. e5 x Ae5 x
8. (9 x  2)e5 x ( Ax  B)e5 x
9. x 2 e5 x ( Ax 2  Bx  C )e5 x
10. e3 x sin 4 x Ae3 x cos 4 x  Be3 x sin 4 x
11. 5 x 2 sin 4 x ( Ax 2  Bx  C ) cos 4 x  ( Ex 2  Fx  G ) sin 4 x
12. xe3 x cos 4 x ( Ax  B )e3 x cos 4 x  (Cx  E )e3 x sin 4 x CH3_18
 Rule of Case 2:

 If any term in yp duplicates a term in yc, it should be

multiplied by xn, where n is the smallest positive integ
er that eliminates that duplication.

CH3_19
If g  x equals a sum?
Suppose that
 The input function g  x  consists of a sum of terms of the kind listed
in the above table i.e.

g  x   g1  x   g 2  x     g m  x .
 The trial forms corresponding to be
g1  x  , g 2  x  ,  , g m  x 
.
y p1 , y p 2 ,  , y p m
Then, the particular solution of the given non-homogeneous
differential equation is

y p  y p1  y p 2    y p m
In other words, the form of is a linear combination of all the
yp
Linearly independent functions generated by repeated differentiation
of the input function .
g (x )
Example 1
Solve the non homogenous equation by using Undetermined Coe
fficient Method.

y  4 y  8 x 2
Solution:
For complementary solution consider homogenous equation


Auxiliary equation yis  4 y  0

m2  4  0
m 2  4
m  2i
yc  e0 x (C1 cos 2 x  C2 sin 2 x)
 For particular solution we assume
y p  Ax 2  Bx  C
y p  2 Ax  B
y p  2 A
Substituting into the given differential equation

y p  4 y p  8x 2
2 A  4( Ax 2  Bx  C )  8 x 2
2 A  4 Ax 2  4 Bx  4C  8 x 2
Equating the coefficients of and constant we have
2
x ,x
x 2 : 4A  8
x : 4B  0
Constant:
 4C  0we get the values of undetermined coefficients
2 Aequations
Solving these
 A  2, B  0, C  1
Thus
yp  2x2  1
Hence the general solution is
y  yc  y p
y  C1 cos 2 x  C2 sin 2 x  2 x 2  1
Example 2
Solve the non homogenous equation by using Undetermined
Coefficient Method.

y  y  sin x
Solution:
Duplication between y p and y c ?

 If a function in the assumed y p is also present in y then this functio

c
n is a solution of the associated homogeneous differential equation.
yp In
this case the obvious assumption for the form of
is not correct.
 In this case we suppose that the input function is made up of n terms
of kinds i.e.
g ( x)  g1 ( x)  g 2 ( x)    g n ( x)
and corresponding to this input function the assumed particular solution

yp
is
y p  y p  y p    y pn
1 2
 If a y pi contain terms that duplicate terms in y c , then y pi that must
be multiplied with x n , n being the least positive integer that elimina
tes the duplication.
Example 3
Find a particular solution of the following non-homogeneous differential
Equation
y //  5 y /  4 y  8e x
Solution:
To find y , we solve the associated homogeneous differential equation
c

y //
 5 y /
 4 y so
We put y  e mx in the given equation, 0 that the auxiliary equation is
 m2 5m4  0  m1, 4

Thus yc  c1ex  c2e4x

x
Since g ( x)  8e

Therefore y p  Ae x
Substituting in the given non-homogeneous differential equation, we
obtain
Ae x  5 Ae x  4 Ae x  8e x
or
0  8e x
Clearly we have made a wrong assumption for y p , as we did not
remove the duplication.
Since Ae is present in yc .Therefore, it is a solution of the associated
x

homogeneous differential equation

y //  5 y /  4 y  0
To avoid this we find a particular solution of the form

y p  Axe x
We notice that there is no duplication between and this new
yc
Assumption for .
yp
Now
y p /  Axe x  Ae x , y p //  Axe x  2 Ae x
Substituting in the given differential equation, we obtain

or Axe x  2 Ae x  5 Axe x  5 Ae x  4 Axe x  8e x .

 3 Ae x  8e x  A   8 3 .
So that a particular solution of the given equation is given by
y p  (8 3)e x
Hence, the general solution of the given equation is

y c  c1e x  c 2 e 4 x  (8 / 3) x e x

Example 4
Find a particular solution of

y //  2 y /  y  e x
Solution:
Consider the associated homogeneous equation

Put y //  2 y /  y  0

y  e mx
Then the auxiliary equation is
m 2  2m  1  (m  1) 2  0
m  1, 1

Roots of the auxiliary equation are real and equal. Therefore,

y c  c1e x  c 2 xe x
Since
g ( x)  e x

Therefore, we assume that

y p  Ae x
This assumption fails because of duplication between and y .
yc p
We multiply with .
x
Therefore, we now assume

y p  Axe x
However, the duplication is still there. Therefore, we again multiply with
x and assume
y p  Ax 2 e x

Since there is no duplication, this is acceptable form of the trial y .

p
1 2 x
yp  x e
2
Example 5
Solve the initial value problem

y //  y  4 x  10 sin x,

Solution y( )  0, y / ( )  2

Associated homogeneous DE
 y //  y  0
Put
y  e mx
Then the auxiliary equation is
m2 1  0  m   i
The roots of the auxiliary equation are complex. Therefore, the
complementary function is

y c  c1 cos x  c 2 sin x
Since

g ( x)  4 x  10 sin x  g1 ( x)  g 2 ( x)
Therefore, we assume that

y p1  Ax  B , y p2  C cosx  D sin x
So that
y p  Ax  B  C cos x  D sin x
Clearly, there is duplication of the functions cos x and sin x .
To remove this duplication we multiply y p 2 with x . Therefore,
we assume that
y p  Ax  B  C x cos x  Dx sin x.

So that yp  2C sin x  Cx cos x  2D cos x  Dx sin x

Substituting into the given non-homogeneous differential
equation,
we have
y p //  y p  Ax  B  2C sin x  2 Dx cos x

Equating constant terms and coefficients of x , sin x ,

x cos x
we obtain

Ax  B  2C sin x  2 Dx cos x  4 x  10 sin x

 B  0, A  4,  2C  10, 2 D  0

So that
A  4, B  0, C  5, D  0
Thus
y p  4 x  5 x cos x
Hence the general solution of the differential equation is

y  yc  y p  c1 cos x  c 2 sin x  4 x - 5 x cos x

We now apply the initial conditions to find c and c

1 2

y ( )  0  c1 cos   c 2 sin   4  5 cos   0

Since
sin   0, cos   1
Therefore
c1  9
Now
y /  9 sin x  c2 cos x  4  5 x sin x  5 cos x

Therefore
y / ( )  2  9 sin   c 2 cos   4  5 sin   5 cos   2

 c 2  7.
Hence the solution of the initial value problem is

y  9 cos x  7 sin x  4 x  5 x cos x.

Example 6
Solve the differential equation

y //  6 y /  9 y  6 x 2  2  12e 3 x
Solution:
The associated homogeneous differential equation is
y //  6 y /  9 y  0
Put
y  e mx

Then the auxiliary equation is

m 2  6m  9  0  m  3, 3
Thus the complementary function is

y c  c1e 3 x  c 2 xe 3 x
Since
g ( x)  ( x 2  2)  12e 3 x  g1 ( x)  g 2 ( x)
We assume that
Corresponding to:
g1 ( x)  x 2  2 y p1  Ax 2  Bx  C
Corresponding to: g 2 ( x)  12e 3x y p 2  De 3 x

Thus the assumed form of the particular solution is

y p  Ax 2  Bx  C  De 3 x
The function e 3 x in y p 2 is duplicated between yc and y p .
Multiplication with x does not remove this duplication. However, if
we multiply y p with x 2 , this duplication is removed.
2
Thus the operative from of a particular solution is

y p  Ax 2  Bx  C  Dx 2 e 3 x
Then
y p  2 Ax  B  2 Dxe3 x  3Dx 2 e 3 x
And
y p  2 A  2 De 3 x  6 Dxe 3 x  6 Dxe3 x  9 Dx 2 e 3 x
Substituting into the given differential equation and collecting like
term, we obtain
 y p //  6 y p /  y p  9 Ax 2  (12 A  9 B ) x  2 A  6 B  9C  2 De 3 x  6 x 2  2  12e 3 x

Equating constant terms and coefficients of x, x 2 and yields

e3x
2 A  6 B  9C  2,  12 A  9 B  0
9 A  6, 2 D  12
Solving these equations, we have the values of the unknown
coefficients

A  2 3, B  8 9 , C  2 3 and D  -6
Thus,
2 8 2
.
y p  x 2  x   6 x 2 e3x
Hence, the general solution
3 9 3

3x 2 2 8 2
y  yc  yp  c1e  c2 xe  x  x   6x2e3x .
3x
3 9 3
Higher –Order Equation
The method of undetermined coefficients can also be used for higher
order equations of the form
dny d n 1 y dy
an  a n 1  ............  a1  a0 y  g ( x)
dx n dx n 1 dx

with constant coefficients. The only requirement is that g (x) consists

of the proper kinds of functions as discussed earlier.
Example 7
Solve y ///  y //  e x cos x
Solution:
To find the complementary function we solve the associated
homogeneous differential equation

y ///  y //  0
 Put
y  e mx , y   me mx , y   m 2 e mx

Then the auxiliary equation is

m3  m 2  0
Or
m 2 ( m  1)  0  m  0,0,1
The auxiliary equation has equal and distinct real roots. Therefore, the
complementary function is

y c  c1  c 2 x  c3 e  x
Since
g ( x)  e x cos x
Therefore, we assume that

y p  Ae x cos x  Be x sin x
Clearly, there is no duplication of terms between and .
yc yp
 Example 8

Solve y"4 y '2 y  2 x 2  3x  6

Solution:
We can get yc as described in Sec 3.3.
Now, we want to find yp.
Since the right side of the DE is a polynomial,
we set
y p  Ax 2  Bx  C , y p , y p '  2 Ax  B, y p " 2 A
After substitution,
2A + 8Ax + 4B – 2Ax2 – 2Bx – 2C = 2x2 – 3x + 6

CH3_44
 Example 8 (2)

Then
 2 A  2 , 8 A  2 B  3 , 2 A  4 B  2C  6
A  1, B  5/2, C  9
2 5
yp  x  x  9
2

CH3_45
 Example 9

Find a particular solution of

y" y ' y  2 sin 3x
Solution:
Let yp = A cos 3x + B sin 3x
After substitution,
(8 A  3B) cos 3x  (3 A  8B) sin 3x  2 sin 3x
Then
A  6/73, B  16/73
6 16
y p  cos 3x  sin 3x
73 73

CH3_46
 Example 10

Solve 6 xe2 x
y"2 y '3 y  4 x  5 (3)
Solution:
x 3x
We can find yc  c1e  c2e
Let y p  Ax  B  Cxe2 x  Ee2 x
After substitution,
2x 2x
 3 Ax  2 A  3B  3Cxe  (2C  3E )e
2x
 4 x  5  6 xe
Then
A  4/3, B  23/9, C  2, E  4/3
4 23 2x 4 2x
y p   x   2 xe  e
3 9 3
x 3x 4 23  4  2x
y  c1e  c2e  x    2 x   e
3 9  3 CH3_47
 Example 11

Find yp of y"5 y '4 y  8e x

Solution:
First let yp = Ae2x
After substitution, 0 = 8e2x, (wrong guess)
Let yp = Axe2x
After substitution, -3Ae2x = 8e2x
Then A = -8/3, yp = (−8/3)xe2x

CH3_48
 Example 12

Find the form of yp of

3 x x
(a) y"8 y '25 y  5 x e  7e
Solution:
3 x
We have g ( x )  (5 x  7
and try) e
y p  ( Ax3  Bx 2  Cx  E )e x
There is no duplication between yp and yc .
(b) y” + 4y = x cos x
Solution:
We try x p  ( Ax  B) cos x  (Cx  E ) sin x
There is also no duplication between yp and yc .

CH3_49
 Example 13

Find the form of yp of

  2
y  9 y  14 y  3x  5 sin 2 x  7 xe 6x

Solution: 2
For 3x2: y p1  Ax  Bx  C

For -5 sin 2x: y p2  E cos 2 x  F sin 2 x

For 7xe :
6x y p3  (Gx  H )e6 x

No term in y p  y p1  yduplicates
p2  y p3 a term in yc

CH3_50
 Example 14

Solve y" y  4 x  10 sin x, y ( )  0, y ' ( )  2

Solution:
yc  c1 cos x  c2 sin x
First trial: yp = Ax + B + C cos x + E sin x (5)
However, duplication occurs. Then we try
yp = Ax + B + Cx cos x + Ex sin x
After substitution and simplification,
A = 4, B = 0, C = -5, E = 0
Then y = c1 cos x + c2 sin x + 4x – 5x cos x
Using y() = 0, y’() = 2, we have
y = 9 cos x + 7 sin x + 4x – 5x cos x
CH3_51
 Example 15
2 3x
Solve y" 6 y '9 y  6 x  2  12 e
Solution:
yc = c1e3x + c2xe3x
y p  Ax2  Bx 
  C  Ee

3x

y p1 y p2

After substitution and simplification,

A = 2/3, B = 8/9, C = 2/3, E = -6
Then
3x 3x 2 2 8 2 2 3x
y  c1e  c2 xe  x  x   6 x e
3 9 3

CH3_52
 Example 16

Solve y   y" e x

cos x
Solution:
m3 + m2 = 0, m = 0, 0, -1
yc = c1+ c2x + c3e-x
yp = Aex cos x + Bex sin x
After substitution and simplification,
A = -1/10, B = 1/5
Then
x 1 x 1 x
y  yc  y p  c1  c2 x  c3e  e cos x  e sin x
10 5

CH3_53
 Example 17

Find the form of yp of

y ( 4)  y  1  x 2e x
Solution:
yc = c1+ c2x + c3x2 + c4e-x
2 x x x
Normal trial: y p  A  Bx e   Cxe    Ee
y p1 y p2

Multiply A by x3 and (Bx2e-x + Cxe-x + Ee-x) by x

Then
yp = Ax3 + Bx3e-x + Cx2e-x + Exe-x

CH3_54
 3.7 Nonlinear Equations
2
Example 1 Solve y "  2 x ( y ' )
Solution:
This nonlinear equation misses y term. Let u(x) = y’,
then du/dx = y”,
duor 2 du
 2xu 2
 2 x dx
dx u
1 (This
2 2
form is just for convenience)
 u  x  c1
dy 1
Since u = 1/y’,
-1
 2 2
dx x  c1
dx 1 1 x
So, y    2 2   tan  c2
x  c1 c1 c1
CH3_55
 Example 2

Solve yy"  ( y ' ) 2

Solution:
This nonlinear equation misses x term. Let u(x) = y’,
then y” = du/dx = (du/dy)(dy/dx) = u du/dy
 du  du dy
y  u or   u 2

 dy  u y
c1
ln|u| = ln|y| + c1, u = c2y (where c2  )e
Since u = dy/dx = c2y, dy/y = c2 dx
ln|y| = c2x + c3,
y  c4ec2 x
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 Example 3

 Assume
y  x  y  y 2 , y(1)
(0)  1 , y(0)  1
exists. If we further assume y(x) possesses a Taylor se
ries centered at 0:
y ( x) (2)
y(0) y(0) 2 y(0) 3 y ( 4) (0) 4 y (5) (0) 5
 y ( 0)  x x  x  x  x 
1! 2! 3! 4! 5!
Remember that y(0) = -1, y’(0) = 1. From the original
DE, y”(0) = 0 + y(0) – y(0)2 = −2.
Then d
y( x)  ( x  y(3)
 y 2 )  1  y  2 yy
dx
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 Example 3 (2)

d
y ( x)  (1  y  2 yy(4)
( 4)
)  y  2 yy  2( y)2
dx

d
y ( x)  ( y  2 yy  2( y(5)
( 5)
)2 )  y  2 yy  6 yy
dx
and so on. So we can use the same method to obtain
y(3)(0) = 4, y(4)(0) = −8, ……
Then
2 2 3 1 4 1 5
y ( x)  1  x  x  x  x  x  
3 3 5

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