Variation of Parameters
We now consider a more general way of solving differential equations of the form
d 2y
dy
a 2 +b
+ cy = f (t )
dt
dt
Suppose we have already found the complementary solution of the related homogeneous
equation is ay ′′ + by ′ + cy = 0 , given by:
yc (t ) = c1y1(t ) + c2y2 (t ) .
In order to find a second solution to the differential equation, this method requires us to
replace the constants (parameters) c1, c2 with unknown functions v1(t ), v2 (t ) . With this
choice our guess for a particular solution will be of the form
y p = v1y1 + v2y2 .
Differentiation yields
) (
(
)
y p′ = v1′y1 + v2′y2 + v1y1′ + v2y2′ .
Since v1(t ) and v2 (t ) are arbitrary, we may impose two conditions on them. First, we will
choose to make y p a solution of the differential equation; the second condition is chosen
to make our work easier. We will choose to let
(v ′y + v ′y ) = 0 .
1
1
2
(0.1)
2
Then
y p′′ = v1′y1′ + v2′y2′ + v1y1′′ + v2y2′′ .
Substituting into the differential equation, we get
)
(
a ⎛⎜ v1′y1′ + v2′y2′ + v1y1′′ + v2y2′′ ⎞⎟ + b v1y1′ + v2y2′ + c (v1y1 + v2y2 ) = f (t )
⎝
⎠
This can be rearranged to show that
(
)
v1 ⎛⎜ ay1′′ + by1′ + cy1 ⎞⎟ + v2 ⎛⎜ ay2′′ + by2′ + cy2 ⎞⎟ + a v1′y1′ + v2′y2′ = f (t ) .
⎝
⎠
⎝
⎠
(0.2)
1
Since y1 and y2 are solutions of the complementary equation, we know that
ay1′′ + by1′ + cy1 = 0 and that ay2′′ + by2′ + cy2 = 0 , so equation (0.2) is reduced to
)
(
a v1′y1′ + v2′y2′ = f (t ) .
(0.3)
The resulting system of equations (0.1)and (0.3)can now be solved for v1′(t ) and v2′(t ) .
We may then be able to find v1(x ) and v2 (x ) directly using integration.
(v ′y + v ′y ) = 0 and a (v ′y ′ + v ′y ′ ) = f (t)
The system of equations
1
1
2
2
1
1
2
2
can be solved using Cramer’s rule (here I assume a = 1 ):
y2
0
y1
0
f (x ) y2′
y1′ f (x )
−y2 f
y1 f
and v2′ = y
v1′ = y y =
=
y
1
2
1
2
W (y1, y2 )
W (y1, y2 )
y1′ y2′
y1′ y2′
Example 1: Using variation of parameters
Solve y ′′ + y = tan x , −
π
2
<x <
π
2
using variation of parameters
Solution: First solve the related homogeneous equation y ′′ + y = 0 . The characteristic
equation is λ 2 + 1 = 0 . Which has roots λ = ±i . Therefore, the complementary solution
is yc = c1 cos x + c2 sin x . With
y1 = cos x and y2 = sin x , we have y1′ = − sin x and
y2′ = cos x . The equations above are then
0
sin x
tan x cos x
− sin x tan x
− sin2 x cos2 x − 1
v1′ =
=
=−
=
= cos x − sec x , and ,
cos x sin x
1
cos x
cos x
v2′ =
− sin x
cos x
cos x
0
sin x
tan x
cos x
sin x
− sin x
cos x
=
cos x tan x
= sin x
1
2
Integrating to solve for v1(x ) and v2 (x ) , we have
∫ v (x )dx = ∫ ( cos x − sec x )dx = sin x − ln sec x + tan x
∫ v (x )dx = ∫ sin xdx = − cos x .
1
2
Therefore,
y p = v1y1 + v2y2 = ( sin x − ln sec x + tan x ) cos x + (− cos x )sin x
= − cos x ln sec x + tan x ,
and the general solution is
y(x ) = yc + y p = c1 cos x + c2 sin x − cos x ln sec x + tan x
Example 2: Using variation of parameters
Solve y ′′ + 9y = 9 sec2 3x , 0 < x <
π
6
using variation of parameters
yc = c1 cos 3x + c2 sin 3x , y p = − sec 3x cos 3x + ln sec 3x + tan 3x sin 3x
3
Example 3: Using variation of parameters
Solve y ′′ + 4y ′ + 4y = x −2e −2x , x > 0 using variation of parameters
y p = −e −2x (ln x − 1) or −e −2x ln x since e −2x appears in yc
4