Variation of Parameters

In the last section we solved nonhomogeneous differential equations using the method of undetermined coefficients.  This method fails to find a solution when the functions g(t) does not generate a UC-Set.  For example if g(t) is sec(t), t -1, ln t, etc, we must use another approach.  The approach that we will use is similar to reduction of order.  Our method will be called variation of parameters.

Consider the differential equation

L(y)  =  y'' + p(t)y' + q(t)y  =  g(t)

And let y1 and y2 be solutions to the corresponding homogeneous differential equation

L(y)  =  0

We write the particular solution is of the form

yp = u1y1 + u2y2

where u1 and u2 are both functions of t.  Notice that this is always possible, by setting

u1  =  1/y1        and        u2  =  (yp - 1)/y2

Actually more can be said, since we are choosing two parameters to find one solution, we can impose one additional condition on the u1 and u2 and still end up with a solution.  We make the assumption that

u1' y1 + u2' y2  =  0

This assumption will come in handy later.

Next take the derivative

yp' = u1'y1 + u1y1' + u2'y2 + u2y2

The assumption helps us simplify yp' as

yp' =  u1y1' +  u2y2

Now take a second derivative

yp'' = u1'y1' + u1y1'' + u2'y2' + u2y2''

Now substitute into the original differential equation to get

(u1'y1' + u1y1'' + u2'y2' + u2y2'') + p(t)(u1y1' +  u2y2' ) + q(t)(u1y1 + u2y2)  =  g(t)

Combine terms with common u's, we get

u1(y1'' + p(t)y1' + q(t)y1) + u2(y2'' + p(t)y2' + q(t)y2) + u1'y1' + u2'y2'  =  g(t)

Now notice that since y1 and y2 are solutions to the differential equation, both expressions in the parentheses are zero.  We have

u1'y1' + u2'y2'  =  g(t)

This equation along with the assumption give a system of two equations and two unknowns

u1' y1 + u2' y2  =  0

u1'y1' + u2'y2'  =  g(t)

Using matrices we get

We recognize the first matrix as the matrix for the Wronskian.  Calling this W, and recalling that the Wronskian of two linearly independent solutions is never zero we can take W-1 of both sides to get

Integrate to find u1 and u2.

Example

Given that

y1  =  x2  and   y2  =  x2 ln x

are solutions to

x2y'' - 3xy' + 4y  =  x2ln x

to the corresponding homogeneous differential equation, find the general solution to the nonhomogeneous differential equation.

Solution

First, we divide by x2 to get the differential equation in standard form

y'' - 3/x y' + 4/x2 y  =  ln x

We let

yp = u1y1 + u2y2

The Wronskian matrix is

We use the adjoint formula to find the inverse matrix.  First the Wronskian is the determinant which is

W  =  x3 + 2x3 ln x - 2x3 ln x  =  x3

So the inverse is

We have

Integrating using u-substitution gives

-(ln x)3
u1  =
3

(ln x)2
u2  =
2

We have

yp  =  -1/3 x2(ln x)3 + 1/2 x2 (ln x)3  =  1/6 (ln x)3

Finally we get

y  =  c1 x2 + c2 x2 ln x + 1/6 (ln x)3