Math 2C
§ 4.6 Variation of Parameters
Introduction: Our goal in this section is to solve linear nonhomogeneous DEs with constant
coefficients, where the input function g ( x ) is not of a form discussed in section 4.4 (linear
combinations of sines, cosines, polynomials and eα x ). Our problem in 4.4 was that we could not find
1
y p when g ( x ) was something of the form , csc x , ln x , etc.
x
The method we examine in this section, due to the astronomer and mathematician Joseph Louis
Lagrange, is known as variation of parameters.
Linear First-Order DEs Revisited:
Recall from section 2.3 Linear First-Order DEs:
dy
dy
+ a0 ( x ) y = g ( x ) →
+ P( x) y = f ( x)
dx
dx
P( x ) dx
Multiplying by the magical function µ x = e ∫
had a magical effect:
a1 ( x )
()
e∫
P( x ) dx
P( x ) dx
P( x ) dx
dy
+ e∫
P ( x ) y = e∫
f ( x)
dx
e∫
P( x ) dx
y = ∫ e∫
→
d ⎡ ∫ P( x ) dx ⎤ ∫ P( x ) dx
e
y⎥ = e
f ( x)
dx ⎢⎣
⎦
Then:
P( x ) dx
f ( x )dx + c
→
− P( x ) dx
− P( x ) dx
P( x ) dx
y = ce ∫
+e ∫
∫ e ∫ f ( x )dx
This has the general form that we learned about in 4.1, y = yc + y p , where
•
− P( x ) dx
dy
+ P( x) y = 0
yc = ce ∫
is a solution of
dx
•
− P( x ) dx
P( x ) dx
dy
yp = e ∫
∫ e ∫ f ( x )dx is a particular solution of dx + P ( x ) y = f ( x )
We now try to find y p a different way, using the method known as variation of parameters. The
procedure is similar to what we did in section 4.2 (reduction of order).
− P( x ) dx
dy
+ P ( x ) y = 0 . It can be verified that y1 = e ∫
is a
dx
solution, and because the equation is first-order, the general solution is y = c1 y1 .
Suppose we know that y1 is a solution of
() ()
Variation of Parameters: Try to find a particular solution y p of the form y p = u1 x y1 x . We have
replaced the constant parameter c1 with a function u1 ( x ) . Recall, we are trying to find u1 so that
y p = u1 ( x ) y1 ( x ) is a solution. Let’s see if we can pull it off.
Zill/Wright – 8e
1
Substitute y p = u1 y1 back into
dy
+ P( x) y = f ( x) :
dx
Linear Second-Order DEs:
Consider the linear second-order equation:
In standard form, this is:
a2 ( x ) y′′ + a1 ( x ) y′ + a0 ( x ) y = g ( x )
y′′ + P ( x ) y′ + Q ( x ) y = f ( x )
(1)
(2)
where P, Q, and f are continuous on some interval I.
As we saw in 4.3, we can find the general solution to the associated homogeneous equation of (2)
when the coefficients are constant; call it yc = c1 y1 + c2 y2 .
We now ask a question similar to above, can we replace the constants c1 and c2 with functions u1 ( x )
and u2 ( x ) so that y p = u1 y1 + u2 y2 is a particular solution of (2)? We need to find those u functions!
Let’s take some derivatives and substitute:
y′p = u1 y1′ + u1′ y1 + u2 y2′ + u2′ y2
y′′p = u1 y1′′+ u1′ y1′ + u1′′y1 + u1′ y1′ + u2 y2′′ + u2′ y2′ + u2′′y2 + u2′ y2′
Zill/Wright – 8e
2
Substituting into (2) gives:
y′′p + P ( x ) y′p + Q ( x ) y p =
= u1 y1′′+ 2u1′ y1′ + u1′′y1 + u2 y2′′ + 2u2′ y2′ + u2′′y2 + P ( x ) ⎡⎣u1 y1′ + u1′ y1 + u2 y2′ + u2′ y2 ⎤⎦ + Q ( x ) ⎡⎣u1 y1 + u2 y2 ⎤⎦
= u1 ⎡⎣ y1′′+ Py1′ + Qy1 ⎤⎦ + u2 ⎡⎣ y2′′ + Py2′ + Qy2 ⎤⎦ + P ⎡⎣u1′ y1 + u2′ y2 ⎤⎦ + u1′′y1 + u1′ y1′ + u2′′y2 + u2′ y2′ + u1′ y1′ + u2′ y2′
Since we are trying to find two unknown functions, we need two equations:
By Cramer’s Rule, the solution of the system is:
u1′ =
W1
W
and u2′ = 2
W
W
where
W=
Then,
u1′ = −
y1
y2
y1′
y2′
y2 f ( x )
W
,
W1 =
0
f ( x)
y2
y2′
u2′ =
and
(
W2 =
,
y1 f ( x )
W
y1
y1′
0
f ( x)
.
)
Note: W is the Wronskian, and we know that W y1 ( x ) , y2 ( x ) ≠ 0 for all x in I because
______________________________ .
Summary of the Method:
Zill/Wright – 8e
3
Example: Solve.
Zill/Wright – 8e
y′′ − 2 y′ + y =
ex
1+ x 2
4
Example: Solve.
Zill/Wright – 8e
2 y′′ + 18y = 6 tan ( 3x )
5