Lecture
14
Variation of Parameters
Variation of Parameters Method for Higher-Order Equations
The method of the variation of parameters just examined for second-order differential
equations can be generalized for an nth-order equation of the type.
an
dny
dx n
 an 1
d n 1 y
dx n 1
   a1
dy
 a0 y  g ( x)
dx
The application of the method to nth order differential equations consists of performing
the following steps.
Step 1 To find the complementary function we solve the associated homogeneous
equation
dny
d n1 y
an n  an1 n1 
dx
dx
 a1
dy
 a0 y  0
dx
Step 2 Suppose that the complementary function for the equation is
y  c1 y1  c2 y 2    cn y n
Then y1 , y 2 ,, y n are n linearly independent solutions of the homogeneous equation.
Therefore, we compute Wronskian of these solutions.
W  y1 , y2 , y3 ,
, yn  
y1
y1
y2
y2
yn
yn
y1( n1)
y2 ( n1)
yn ( n1)
Step 4 We write the differential equation in the form
y   Pn1  x  y
n
n1

and compute the determinants Wk ; k  1, 2,
0
 P1  x  y  P  x  y  f  x 
, n ; by replacing the kth column of W by
0
the column

0
f ( x)
Step 5 Next we find the derivatives u1 , u 2 , , u n of the unknown functions u1, u2 ,, un
through the relations
1
Lecture
14
Variation of Parameters
Wk
,
k  1, 2, , n
W
Note that these derivatives can be found by solving the n equations
u k 
y1u1
y1 u1



y 2u 2
y 2 u 2
   y n u n
   y n u n


0

0

y1n 1u1  y 2 n 1u 2    y n n 1u n  f  x 
Step 6 Integrate the derivative functions computed in the step 5 to find the functions u k
Wk
uk  
dx,
k  1, 2, , n

W
Step 7 We write a particular solution of the given non-homogeneous equation as
y p  u1  x  y1  x   u2  x  y2  x  
 un  x  yn  x 
Step 8 Having found the complementary function y c and the particular integral y p , we
write the general solution by substitution in the expression
y  yc  y p
Note that
 The first n  1 equations in step 5 are assumptions made to simplify the first
n  1 derivatives of y p . The last equation in the system results from substituting
the particular integral y p and its derivatives into the given nth order linear
differential equation and then simplifying.

Depending upon how the integrals of the derivatives u k of the unknown
functions are found, the answer for y p may be different for different attempts to
find y p for the same equation.

When asked to solve an initial value problem, we need to be sure to apply the
initial conditions to the general solution and not to the complementary function
alone, thinking that it is only y c that involves the arbitrary constants.
Example 1
2
Lecture
14
Variation of Parameters
Solve the differential equation by variation of parameters.
d3y
dx3

dy
 csc x
dx
Solution
Step 1: The associated homogeneous equation is
d3y
dx
3

dy
0
dx
Auxiliary equation


m3  m  0  m m 2  1  0
m  0, m   i
Therefore the complementary function is
y  c1  c2 cos x  c3 sin x
c
Step 2: Since
y  c1  c2 cos x  c3 sin x
c
Therefore
y1  1,
y2  cos x,
y3  sin x
So that the Wronskian of the solutions y1 , y 2 and y3
1 cos x
sin x
W  y1 , y2 , y3   0  sin x cos x
0  cos x  sin x
By the elementary row operation R1  R3 , we have
1
0
0

0
0
 sin x
 cos x
cos x
 sin x

 sin 2 x  cos2 x  1  0
Step 3: The given differential equation is already in the required standard form
y  0 y  y  0 y  csc x
3
Lecture
14
Variation of Parameters
Step 4: Next we find the determinants W1 ,W2 and W3 by respectively, replacing 1st, 2nd
0
and 3rd column of W by the column 0
csc x
0
cos x
sin x
W1  0
 sin x cos x
csc x  cos x  sin x


 csc x sin 2 x  cos 2 x  csc x
1
0
sin x
W2  0
0
cos x
0 csc x  sin x

and
0
cos x
  cos x csc x   cot x
csc x  sin x
1 cos x
0
 sin x
0
  sin x csc x   1
W3  0  sin x
0 
 cos x csc x
0  cos x csc x
Step 5: We compute the derivatives of the functions u1 , u 2 and u3 as:
u1 
W1
 csc x
W
u 2 
W2
  cot x
W
u 3 
W3
 1
W
Step 6: Integrate these derivatives to find u1 , u 2 and u3
W1
u1  
dx   csc xdx  ln csc x  cot x

W
W
  cos x
u2   2 dx    cot xdx  
dx   ln sin x
 sin x
 W
4
Lecture
14
Variation of Parameters
W3
u3  
dx    1dx   x

W
Step 7: A particular solution of the non-homogeneous equation is
y  ln csc x  cot x  cos x ln sin x  x sin x
p
Step 8: The general solution of the given differential equation is:
y  c1  c2 cos x  c3 sin x  ln csc x  cot x  cos x ln sin x  x sin x
Example 2
Solve the differential equation by variation of parameters.
y   y   tan x
Solution
Step 1: We find the complementary function by solving the associated homogeneous
equation
y   y   0
Corresponding auxiliary equation is


m3  m  0  m m 2  1  0
m  0, m   i
Therefore the complementary function is
yc  c1  c2 cos x  c3 sin x
Step 2: Since
yc  c1  c2 cos x  c3 sin x
Therefore
y1  1,
y2  cos x,
y3  sin x
Now we compute the Wronskian of y1 , y 2 and y3
1 cos x
sin x
W  y1 , y2 , y3   0  sin x cos x
0  cos x  sin x
By the elementary row operation R1  R3 , we have
1
0
0
0
0
 sin x
 cos x
cos x
 sin x
5
Lecture
14
Variation of Parameters


 sin 2 x  cos2 x  1  0
Step 3: The given differential equation is already in the required standard form
y   0  y   y   0  y  tan x
Step 4: The determinants W1 ,W2 and W3 are found by replacing the 1st, 2nd and 3rd
column of W by the column
0
0
tan x
Therefore
0
cos x
sin x
W1  0
 sin x cos x
tan x  cos x  sin x


 tan x cos 2 x  sin 2 x  tan x
1
0
sin x
W2  0
0
cos x  10  cos x tan x   sin x
0 tan x  sin x
1
and
W3  0
0
cos x
0
 sin x
 cos x
0  1 sin x tan x  0   sin x tan x
tan x
Step 5: We compute the derivatives of the functions u1 , u 2 and u3 .
W
u1  1  tan x
W
u2 
W2
  sin x
W
u3 
W3
  sin x tan x
W
Step 6: We integrate these derivatives to find u1 , u 2 and u3
W
1
u1  

W
W
2
u2  

W

dx   tan x dx   

sin x
dx   ln cos x
cos x
dx    sin x dx  cos x
6
Lecture
14
Variation of Parameters
W
3
u3  

W

dx    sin x tan xdx
   sin x


sin x
dx    sin 2 x sec dx
cos x



  cos 2 x  1 sec xdx   cos 2 x sec x  sec x dx
   cos x  sec x  dx   cos xdx   sec xdx
 sin x  ln sec x  tan x
Step 7: Thus, a particular solution of the non-homogeneous equation
y   ln cos x  cos x cos x   sin x  ln sec x  tan x
p
  sin x 
  ln cos x  cos 2 x  sin 2 x  sin x ln sec x  tan x
  ln cos x  1  sin x ln sec x  tan x
Step 8: Hence, the general solution of the given differential equation is:
y  c1  c2 cos x  c3 sin x  ln cos x  1  sin x ln sec x  tan x
or
y  c1  1  c2 cos x  c3 sin x  ln cos x  sin x ln sec x  tan x
or
y  d1  c2 cos x  c3 sin x  ln cos x  sin x ln sec x  tan x
where d1 represents c1  1 .
Example 3
Solve the differential equation by variation of parameters.
y  2 y  y  2 y  e3x
Solution
Step 1: The associated homogeneous equation is
y  2 y  y  2 y  0
The auxiliary equation of the homogeneous differential equation is
m3  2m2  m  2  0


 (m  2) m2  1  0
 m  1, 2, 1
The roots of the auxiliary equation are real and distinct. Therefore yc is given by
yc  c1e x  c2e2 x  c3e  x
Step 2: From yc we find that three linearly independent solutions of the homogeneous
differential equation.
7
Lecture
14
Variation of Parameters
y1  e x , y2  e2 x , y3  e x
Thus the Wronskian of the solutions y1 , y 2 and y3 is given by
e x e2 x
e x
1 1 1
x
2
x

x
x
2
x

x
We
2e
e
 e  e  e 1 2 1
1 4 1
e x 4e2 x e x
By applying the row operations R2  R1, R3  R1 , we obtain
1 1
1
W  e2 x 0 1 2  6e2 x  0
0 3 0
Step 3: The given differential equation is already in the required standard form
y  2 y  y  2 y  e3x
Step 4: Next we find the determinants W1 ,W2 and W3 by, respectively, replacing the 1st,
2nd and 3rd column of W by the column
0
0
e3x
Thus
0
e2 x
W1  0
2e2 x
e3x
4e2 x
e x
2x
31 e

x
e
  1
2e2 x

x
e
e x
e  x
e3x


 e3x e x  2e x  3e4 x
ex
0
W2  e x
ex
0

e3x
e x
x
3 2 e

x
e
  1
ex

x
e
e x
e  x
e3x

  e0  e0 e3x  2e3x
8
Lecture
and
14
Variation of Parameters
ex
e2 x
W3  e x
2e2 x
ex
4e2 x
0
ex
3
x
0 e
ex
e3x
e2 x
2e2 x


 e3x 2e3x  e3x  e6 x
Step 5: Therefore, the derivatives of the unknown functions u1 , u 2 and u3 are given by.
u1 
W1  3e 4 x
1

  e2x
2
x
W
2
6e
W2 2e3x 1 x

 e
W 6 e2 x 3
W3
e6x
1
u3 
 2x  e4x
W
6
6e
Step 6: Integrate these derivatives to find u1 , u 2 and u3
u2 
W1
1 2x
1 2x
1 2x
u1  
dx  
  e dx    e dx   e

 2
2
4
W
W
1
1
u2   2 dx   e x dx  e x
3
3
 W
W3
1 4x
1 4x
u3  
dx  
e
 e dx 

6
24
W
Step 7: A particular solution of the non-homogeneous equation is
1
1
1
y p   e3 x  e3 x  e3 x
4
3
24
Step 8: The general solution of the given differential equation is:
1
1
1
y  c1e x  c2e2 x  c3e x  e3x  e3x  e3x
4
3
24
9
Lecture
14
Variation of Parameters
Practice Exercise
Solve the higher order differential equations by variations of parameters.
1. y   4 y   sec 2 x
2. 2 y   6 y   x 2
Solve the initial value problems.
3. 2 y   y   y  x  1


4. y   4 y   4 y  12 x 2  6 x e 2 x
10
Lecture
14
Variation of Parameters
Differential Equations with Variable Coefficients
So far we have been solving Linear Differential Equations with constant coefficients.
We will now discuss the Differential Equations with non-constant (variable) coefficients.
These equations normally arise in applications such as temperature or potential u in the
region bounded between two concentric spheres. Then under some circumstances we
have to solve the differential equation:
d 2u
du
r 2 2
0
dr
dr
where the variable r>0 represents the radial distance measured outward from the center
of the spheres.
Differential equations with variable coefficients such as
x 2 y   xy   ( x 2  v 2 ) y  0
(1  x 2 ) y   2 xy   n(n  1) y  0
y   2 xy   2ny  0
and
occur in applications ranging from potential problems, temperature distributions and
vibration phenomena to quantum mechanics.
The differential equations with variable coefficients cannot be solved so easily.
Cauchy- Euler Equation:
Any linear differential equation of the form
n 1
dny
y
d y
n 1 d

a
x
   a1 x
 a 0 y  g ( x)
n 1
n
n 1
dx
dx
dx
where an , an1 ,, a0 are constants, is said to be a Cauchy-Euler equation or equidimensional equation. The degree of each monomial coefficient matches the order of
differentiation i.e x n is the coefficient of nth derivative of y, x n 1 of (n-1)th derivative of
y, etc.
an x n
For convenience we consider a homogeneous second-order differential equation
d2y
dy
 cy  0, x  0
dx
dx 2
The solution of higher-order equations follows analogously.
ax 2
 bx
11
Lecture
14
Variation of Parameters
Also, we can solve the non-homogeneous equation
2
dy
2d y
ax
 bx  cy  g ( x),
2
dx
x0
dx
by variation of parameters after finding the complementary function y c (x).
We find the general solution on the interval (0, ) and the solution on (0,) can be
obtained by substituting t   x in the differential equation.
Method of Solution:
We try a solution of the form y  x m , where m is to be determined. The first and second
derivatives are, respectively,
dy
d2y
 mx m 1 and
 m(m  1) x m  2
2
dx
dx
Consequently the differential equation becomes
d2y
dy
ax 2 2  bx  cy  ax 2  m(m  1) x m2  bx  mx m1  cx m
dx
dx
 am(m  1) x m  bmx m  cx m
 xm (am(m  1)  bm  c)
Thus y  x m is a solution of the differential equation whenever m is a solution of the
auxiliary equation
(am(m  1)  bm  c)  0 or am 2  (b  a)m  c  0
The solution of the differential equation depends on the roots of the AE.
Case-I: Distinct Real Roots
Let m1 and m2 denote the real roots of the auxiliary equation such that m1  m2 . Then
y  x m1 and y  x m2 form a fundamental set of solutions.
Hence the general solution is
y  c1 x m1  c2 x m2 .
12
Lecture
14
Variation of Parameters
Example 1
Solve
2
dy
2d y
x
 2x  4 y  0
2
dx
dx
Solution:
Suppose that y  x m , then
dy
d2y
 mx m 1 ,
 m(m  1) x m  2
2
dx
dx
Now substituting in the differential equation, we get:
2
dy
2 d y
x
 2 x  4 y  x 2  m(m  1) x m2  2 x  mx m1  4 x m
2
dx
dx
 x m (m(m  1)  2m  4)
x m (m2  3m  4)  0 if m 2  3m  4  0
This implies m1  1, m2  4 ; roots are real and distinct.
So the solution is
y  c1 x 1  c2 x 4 .
Case II: Repeated Real Roots
If the roots of the auxiliary equation are repeated, that is, then we obtain only one
solution y  x m1 .
To construct a second solution y 2 , we first write the Cauchy-Euler equation in the form
d 2 y b dy
c

 2 y0
2
ax dx ax
dx
Comparing with
d2y
dy
 P( x)  Q( x) y  0
2
dx
dx
b
We make the identification P ( x ) 
. Thus
ax
b
 dx
e ax
m1
y 2  x  m1 2 dx
(x )
 x m1 
e
b
 ( ) ln x
a
x 2 m1
dx
13
Lecture
14
Variation of Parameters

x
xm1
b
a
.x 2m1 dx
Since roots of the AE am 2  (b  a)m  c  0 are equal, therefore discriminant is zero
(b  a )
(b  a )
i.e m1  
or 2m1  
2a
a
y2 
b b  a
x a .x a dx

dx
 x x
x m1
y 2  x m1
m1
ln x.
The general solution is then
y  c1 x m1  c2 x m1 ln x
Example 2
4x 2
Solve
d2y
dy
 8 x  y  0.
2
dx
dx
Solution:
Suppose that y  x m , then
dy
d2y
m 1
 mx
,
 m(m  1) x m2 .
2
dx
dx
Substituting in the differential equation, we get:
d2y
dy
4 x 2 2  8 x  y  x m (4m(m  1)  8m  1)  x m (4m 2  4m  1)  0
dx
dx
if 4m 2  4m  1  0 or (2m  1) 2  0 .
1
Since m1   , the general solution is
2
y  c1 x

1
2
 c2 x

1
2
ln x .
For higher order equations, if m1 is a root of multiplicity k, then it can be shown that:
xm1 , xm1 ln x, xm1 (ln x)2 , , x m1 (ln x)k 1 are k linearly independent solutions.
Correspondingly, the general solution of the differential equation must then contain a
linear combination of these k solutions.
Case III Conjugate Complex Roots
If the roots of the auxiliary equation are the conjugate pair
14
Lecture
14
Variation of Parameters
m1    i , m2    i
where  and  >0 are real, then the solution is
y  c1x i   c2 x i  .
But, as in the case of equations with constant coefficients, when the roots of the auxiliary
equation are complex, we wish to write the solution in terms of real functions only. We
note the identity
xi   (eln x )i   ei  ln x ,
which, by Euler’s formula, is the same as
xi   cos(  ln x)  i sin(  ln x)
Similarly we have
xi   cos(  ln x)  i sin(  ln x)
Adding and subtracting last two results yields, respectively,
xi   x i   2cos(  ln x)
and
xi   xi   2i sin(  ln x)
From the fact that y  c1x i   c2 x i  is the solution of ax 2 y  bxy  cy  0 ,
for any values of constants c1 and c2 , we see that
y1  x ( xi   x i  ), (c1  c2  1)
y2  x ( xi   x i  ), (c1  1, c2  1)
or
y1  2 x (cos(  ln x))
y2  2 x (sin(  ln x))
are also solutions.
Since W ( x cos(  ln x), x sin(  ln x))  x 2 1  0;   0 , on the interval (0, ), we
conclude that
y1  x cos(  ln x) and y 2  x sin(  ln x)
constitute a fundamental set of real solutions of the differential equation.
Hence the general solution is
y1  x [c1 cos(  ln x)  c2 sin(  ln x)]
Example 3
Solve the initial value problem
d2y
dy
y (1)  1, y (1)  5
x 2 2  3x  3 y  0,
dx
dx
Solution:
15
Lecture
14
Variation of Parameters
Let us suppose that:
y  x m , then
dy
 mx m 1 and
dx
d2y
 m(m  1) x m2 .
2
dx
d2y
dy
 3x  3 y  x m (m(m  1)  3m  3)  x m (m 2  2m  3)  0
2
dx
dx
2
if m  2m  3  0 .
x2
From the quadratic formula we find that m1  1  2i and m1  1  2i . If we make
the identifications   1 and   2 , so the general solution of the differential
equation is
y1  x 1 [c1 cos( 2 ln x)  c 2 sin( 2 ln x)] .
By applying the conditions y (1)  1, y (1)  5 , we find that
c1  1
and c 2  2 2 .
Thus the solution to the initial value problem is
y1  x 1 [cos( 2 ln x)  2 2 sin( 2 ln x)]
Example 4
Solve the third-order Cauchy-Euler differential equation
d3y
d2y
dy
x 3 3  5 x 2 2  7 x  8 y  0,
dx
dx
dx
Solution
The first three derivative of y  x m are
dy
d3y
d2y
m2
 mx m 1 ,
 m(m  1)(m  2) x m 3 ,

m
(
m

1
)
x
,
2
3
dx
dx
dx
so the given differential equation becomes
x3
2
d3y
dy
2 d y

5
x
 7 x  8 y  x 3 m(m  1)( m  2) x m3  5 x 2 m(m  1) x m2  7 xmxm1  8 x m ,
3
2
dx
dx
dx
 x m (m(m  1)( m  2)  5m(m  1)  7m  8)
 x m (m 3  2m 2  4m  8)
In this case we see that y  x m is a solution of the differential equation, provided m is a
root of the cubic equation
m 3  2m 2  4m  8  0
or (m  2)(m 2  4)  0
The roots are: m1  2, m2  2i, m3  2i .
16
Lecture
14
Variation of Parameters
Hence the general solution is
y1  c1 x 2  c 2 cos( 2 ln x)  c3 sin( 2 ln x)
Example 5
Solve the non-homogeneous equation
x2 y  3xy  3 y  2 x 4e x
Solution
Put y  x m
dy
d2y
 mx m 1 ,
 m(m  1) x m2
2
dx
dx
Therefore we get the auxiliary equation,
m(m  1)  3m  3  0 or (m  1)( m  3)  0 or m  1,3
y c  c1 x  c 2 x 3
Thus
Before using variation of parameters to find the particular solution y p  u1 y1  u 2 y 2 ,
recall that the formulas u1 
W2 
y1
y1
0
W1
W
and u 2  2 , where W1 
f ( x)
W
W
y2
,
y 2
0
, and W is the Wronskian of y1 and y 2 , were derived under the
f ( x)
assumption that the differential equation has been put into special form .
y   P( x) y   Q( x) y  f ( x)
3
3
Therefore we divide the given equation by x 2 , and form y   y   2 y  2 x 2 e x
x
x
2 x
we make the identification f ( x)  2 x e . Now with y1  x , y 2  x 2 , and
x
x
x x3
0
x3
3
 2x 3e x
,
W 

2
x
W

 2 x 5 e x , W2 
1
2 x
2
2 x
2
1
2
x
e
1 3x
2 x e 3x
we find
2x 5e x
2x 3e x
2 x



x
e
u

 ex
and
2
3
3
2x
2x
u1   x 2 e x  2 xex  2e x and u 2  e x .
u 1 
17
Lecture
14
Variation of Parameters
y p  u1 y1  u 2 y 2
Hence
 ( x 2 e x  2 xex  2e x ) x  e x x 3  2 x 2 e x  2 xex
Finally we have y  yc  y p  c1 x  c2 x 3  2 x 2 e x  2 xex
Practice Exercise
2.
4 x2 y  y  0
xy  y  0
3.
x 2 y  5 xy  3 y  0
4.
4 x2 y  4 xy  y  0
5.
x 2 y  7 xy  41y  0
6.
x3
1.
8.
2
d3y
2 d y  4 x dy  4 y  0

2
x
dx
dx3
dx2
d4y
d3y
d2y
dy
x4 4  6 x3 3  9 x2 2  3x  y  0
dx
dx
dx
dx
x2 y  5xy  8 y  0; y(1)  0, y(1)  4
9.
x2 y  2 xy  2 y  x3 ln x
10.
d3y
d2y
dy
3
2
x
 3x
 6 x  6 y  3  ln x3
3
2
dx
dx
dx
7.
18