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 UCSet. 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 y_{1} and y_{2} be solutions to the corresponding homogeneous differential equation
L(y) = 0
We write the particular solution is of the form
y_{p} = u_{1}y_{1} + u_{2}y_{2}
where u_{1} and u_{2} are both functions of t. Notice that this is always possible, by setting
u_{1} = 1/y_{1 } and u_{2} = (y_{p}  1)/y_{2}
Actually more can be said, since we are choosing two parameters to find one solution, we can impose one additional condition on the u_{1} and u_{2} and still end up with a solution. We make the assumption that
u_{1}' y_{1} + u_{2}' y_{2} = 0
This assumption will come in handy later.
Next take the derivative
y_{p}' = u_{1}'y_{1} + u_{1}y_{1}' + u_{2}'y_{2} + u_{2}y_{2}'
The assumption helps us simplify y_{p}' as
y_{p}' = u_{1}y_{1}' + u_{2}y_{2}'
Now take a second derivative
y_{p}'' = u_{1}'y_{1}' + u_{1}y_{1}'' + u_{2}'y_{2}' + u_{2}y_{2}''
Now substitute into the original differential equation to get
(u_{1}'y_{1}' + u_{1}y_{1}'' + u_{2}'y_{2}' + u_{2}y_{2}'') + p(t)(u_{1}y_{1}' + u_{2}y_{2}' ) + q(t)(u_{1}y_{1} + u_{2}y_{2}) = g(t)
Combine terms with common u's, we get
u_{1}(y_{1}'' + p(t)y_{1}' + q(t)y_{1}) + u_{2}(y_{2}'' + p(t)y_{2}' + q(t)y_{2}) + u_{1}'y_{1}' + u_{2}'y_{2}' = g(t)
Now notice that since y_{1} and y_{2} are solutions to the differential equation, both expressions in the parentheses are zero. We have
u_{1}'y_{1}' + u_{2}'y_{2}' = g(t)
This equation along with the assumption give a system of two equations and two unknowns
u_{1}' y_{1} + u_{2}' y_{2} = 0
u_{1}'y_{1}' + u_{2}'y_{2}' = 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 u_{1} and u_{2}.

Example
Given that
y_{1} = x^{2} and y_{2} = x^{2 }ln x
are solutions to
x^{2}y''  3xy' + 4y = x^{2}ln x
to the corresponding homogeneous differential equation, find the general solution to the nonhomogeneous differential equation.
Solution
First, we divide by x^{2} to get the differential equation in standard form
y''  3/x y' + 4/x^{2} y = ln x
We let
y_{p} = u_{1}y_{1} + u_{2}y_{2}
The Wronskian matrix is
We use the adjoint formula to find the inverse matrix. First the Wronskian is the determinant which is
W = x^{3} + 2x^{3} ln x  2x^{3} ln x = x^{3}
So the inverse is
We have
Integrating using usubstitution gives
(ln x)^{3}
u_{1}
=
3
(ln x)^{2}
u_{2}
=
2
We have
y_{p} = 1/3 x^{2}(ln
x)^{3} + 1/2 x^{2} (ln x)^{3} = 1/6 x^{2}(ln x)^{3}
Finally we get
y =
c_{1} x^{2} + c_{2} x^{2} ln x + 1/6 x^{2} (ln x)^{3}
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