M ATH 234 - VARIATION OF PARAMETERS
In the last lecture, we saw how to solve the equation
ay 00 + by 0 + cy = g(x),
in the case when g(x) is a polynomial combination of exponentials, sines, cosines, and polynomials. Although
these cases are very important for applications, we would still like to know what to do when g(x) is not of
this form. In this lecture, we will see a general method to answer this question. We will do even better: the
method well use gives us a particular solution to any equation of the form
y 00 + p(x)y 0 + q(x)y = g(x),
even when p(x) and q(x) are not constant.
To do this, well need two linearly independent solutions y1 and y2 of the homogeneous problem
y 00 + p(x)y 0 + q(x)y = 0.
We know that to find the general solution, all we need is the solution of this homogeneous problem, and one
particular solution yp. Then the general solution is given by
y = c 1 y1 + c 2 y2 + y p .
Since we have already seen how to get y2 if we know y1 (using Abels formula, reduction of order), this in
e↵ect means that after today’s lecture we will be able to solve any linear second order equation, as long as
we manage to somehow find one solution of the homogeneous problem.
VARIATION OF PARAMETERS
This is a general method to construct one yp , given y1 and y2 . It works for any g(x). You should realize
that when the method of undetermined coefficients can be used, it is a lot easier.
The general solution to the homogeneous problem is
y = c 1 y1 + c 2 y2 ,
where c1 and c2 are constants. We know that if we plug this in to the whole equation (with g(x)), then the
left-hand side vanishes. That’s not quite what we want, but it’s not altogether terrible: we do want a lot of
things to cancel.
What would happen if c1 and c2 were not constant, but were allowed to depend on x? This is the premise
of the method of variation of parameters: maybe we can fabricate a new solution out of an old solution to
a related problem, by letting whatever parameters are present in that solution vary. Here those parameters
are c1 and c2 .
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So, let’s try to construct a particular solution yp = u1 (x)y1 (x) + u2 (x)y2 (x).
So, we are assuming that yp has a form similar to the homogeneous solution, but a little more complicated.
One thing to note before we proceed: we are trying to find one particular solution, and we have introduced
two new unknown functions u1 and u2 . This means that we still get to introduce one constraint between
these two functions. We will do this when it will be convenient, see below.
Let’s substitute our form of the particular solution into the equation. To that end, we need yp0 and yp00 . So,
let’s take a derivative:
yp0 = u01 y1 + u1 y10 + u02 y2 + u2 y20
Our original equation for yp0 is of second order. If we will take another derivative of yp0 , we will see that
yp00 depends on u001 and u002 . Thus, the di↵erential equation we will have to solve for u1 and u2 will also be
of second order. That is not any simpler than the problem we started with. This is where we will impose
another condition on u1 and u2 . We impose that
u01 y1 + u02 y2 = 0
Then
yp0 = u1 y10 + u2 y20 ,
and yp00 will not depend on second derivatives of u1 or u2 :
yp00 = u1 y100 + u2 y200 + u01 y10 + u02 y20 .
Now we plug everything into our di↵erential equation:
u1 y100
yP00 + pyP0 + qyP = g
00
0 0
0 0
0
0
+ u2 y2 + u1 y1 + u2 y2 + p (u1 y1 + u2 y2 ) + q (u1 y1 + u2 y2 ) = g
u1 (y100 + py10 + qy1 ) + u2 (y200 + py20 + qy2 ) + u01 y10 + u02 y20 = g,
but y1 and y2 are solution of
so that
y 00 + py 0 + qy = 0,
u01 y10 + u02 y20 = g.
Thus the two equations we have to solve for u1 and u2 are
u01 y1 + u02 y2
=
0
u01 y10
=
g
+
u02 y20
We’ll solve these two linear algebraic equations for u01 and u02 , then we’ll integrate to get u1 and u2 .
First, let’s multiply the first equation by y10 and the second equation by y1 . Subtracting the two resulting
equations from each other, the terms with u01 drop out and we get
u02 (y1 y20
y2 y10 )
0
u2 W (y1 , y2 )
gy1
u02 =
W (y1 , y2 )
Z
gy1
u2 =
dx.
W (y1 , y2 )
2
= gy1
= gy1
Note that y1 and y2 are by definition linearly independent, so that W (y1 , y2 ) 6= 0.
Second, let’s multiply the first equation by y20 and the second equation by y2 . Subtracting the two resulting
equations from each other (sounds familiar?), the terms with u02 drop out and we get
Z
gy2
u1 =
dx,
W (y1 , y2 )
proceeding in an identical way to before. In both of these integrals, we don’t care about the constants
of integration since we just want to find one particular solution. As we’ve seen before, finding a di↵erent
particular solution merely results in di↵erent constants c1 and c2 in the form of the general solution. Using
yP = u1 y1 + u2 y2 , we get
yP =
y1
Z
y2 g
dx + y2
W (y1 , y2 )
Z
y1 g
dx.
W (y1 , y2 )
This is the most general form of the particular solution, for any given g(x).
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Example
Find the general solution to the di↵erential equation
y 00
5y 0 + 6y = 2ex .
Solution
Here g(x) = 2ex . Notice that this is an example where we could use either the method of undetermined
coefficients or our new method of variation of parameters. No matter which method we use, we first have to
solve the homogeneous problem.
Homogeneous solution: the characteristic equation is
2
from which
1
= 3 and
2
5 + 6 = 0,
= 2. Thus
y1 = e3x ,
y2 = e2x ,
and
yH = c1 e3x + c2 e2x .
Particular solution, using variation of parameters: first we compute the Wronskian:
W (y1 , y2 ) = y1 y20
y10 y2 = e3x 2e2x
3e3x e2x =
Then
yp = y1 u 1 + y2 u 2
Z
Z
y2 g
y1 g
= y1
dx + y2
dx
W (y1 , y2 )
W (y1 , y2 )
Z 2x x
Z 3x x
e 2e
e 2e
= e3x
dx + e2x
dx
e5x
e5x
Z
Z
= 2e3x e 2x dx 2e2x e x dx
= 2e3x
=
e
2x
2
ex + 2ex
2e2x
e
x
1
= ex .
General solution: the general solution is
= c1 e3x + c2 e2x + ex .
4
e5x .
Example
Find the general solution to the di↵erential equation
y 00 + 9y = 9 sec2 3x.
Solution
This example can’t be done using the method of undetermined coefficients so variation of parameters is our
only option.
Homogeneous solution: the characteristic equation is
2
+ 9 = 0 =)
1,2
= ±3i,
and
y1 = cos 3x, y2 = sin 3x.
The homogeneous solution is
yH = c1 cos 3x + c2 sin 3x.
Particular solution, using variation of parameters: first we compute the Wronskian:
W (y1 , y2 ) = y1 y20
y10 y2 = (cos 3x)(3 cos 3x)
( 3 sin 3x)(sin 3x) = 3.
Then (using u = 3x and v = cos u)
Z
yp = y1
Z
y2 g
y1 g
dx + y2
dx
W (y1 , y2 )
W (y1 , y2 )
Z
Z
9 sec2 3x
9 sec2 3x
= cos 3x
sin 3xdx + sin 3x
cos 3xdx
3
3
Z
Z
cos 3x sec2 u sin udu + sin 3x sec2 u cos udu
Z
Z
1
cos 3x
dv + sin 3x sec udu
v2
1
= cos 3x + sin 3x ln | sec u + tan u|
v
1
cos 3x
+ sin 3x ln | sec u + tan u|
cos u
1
cos 3x
+ sin 3x ln | sec 3x + tan 3x|
cos 3x
= 1 + sin 3x ln | sec 3x + tan 3x|.
General solution: the general solution is given by
y = c1 cos 3x + c2 sin 3x
1 + sin 3x ln | sec 3x + tan 3x|.
As a final note, we should be careful when we start the method of variation of parameters: the form of the
equation for which our solution formula is valid requires that the coefficient of y 00 is one. Thus, if there is
any other coefficient there originally, we have to divide the equation by it, so we can read of the correct form
of g(x).
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