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# 3.5: Variation of Parameters

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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 $$y_1$$ and $$y_2$$ be solutions to the corresponding homogeneous differential equation

$L(y) = 0.$

We write the particular solution is of the form

$y_p = u_1y_1 + u_2y_2$

where $$u_1$$ and $$u_2$$ are both functions of $$t$$. Notice that this is always possible, by setting

$u_1 = \dfrac {1}{y_1} \;\;\; \text{and} \;\;\; u_2 = \dfrac {( y_p - 1)}{y_2}.$

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

$y'_p = u'_1y_1 + u_1y'_1 + u'_2y_2 + u_2y'_2.$

The assumption helps us simplify $$y'_p$$ as

$y'_p = u_1y'_1 + u_2y'_2.$

Now take a second derivative

$y''_p = u'_1y'_1 + u_1y''_1 + u'_2y'_2 + u_2y''_2.$

Now substitute into the original differential equation to get

$(u'_1y'_1 + u_1y''_1 + u'_2y'_2 + u_2y''_2) + p(t)(u_1y'_1 + u_2y'_2) + q(t)(u_1y_1 + u_2y_2) = g(t).$

Combine terms with common $$u$$'s, we get

$u_1 (y''_1 + p(t)y'_1 + q(t)y_1) + u_2(y''_2 + p(t)y'_2 + q(t)y_2) + u'_1y'_1 + u'_2y'_2 = g(t).$

Now notice that since $$y_1$$ and $$y_2$$ are solutions to the differential equation, both expressions in the parentheses are zero. We have

$u'_1y'_1 + u'_2y'_2 = g(t).$

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

$u'_1y_1 + u'_2y_2 = 0$

$u'_1y'_1 + u'_2y'_2 = g(t).$

Using matrices we get

$\begin{pmatrix} y_1 y_2 \\ y'_1 y'_2 \end{pmatrix} \begin{pmatrix} u'_1\\ u'_2 \end{pmatrix} = \begin{pmatrix} 0 \\ g(t) \end{pmatrix}$

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

$\begin{pmatrix} u'_1\\ u'_2 \end{pmatrix} = W^{-1} \begin{pmatrix} 0 \\ g(t) \end{pmatrix}$

We integrate to find $$u_1$$ and $$u_2$$.

$\begin{pmatrix} u_1 \\ u_2 \end{pmatrix} = \int W^{-1} \begin{pmatrix} 0 \\ g(t) \end{pmatrix} dt.$

Example $$\PageIndex{1}$$: Solving a nonhomogeneous differential equation

Given that

$y_1 = x^2 \quad \text{ and } \quad y_2 = x^2 \ln x \nonumber$

are solutions to

$x^2y'' - 3xy' + 4y = x^2 \ln x \nonumber$

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

$y'' - \dfrac {3}{x} y' + \dfrac {4}{x^2}y = \ln x.\nonumber$

We let

$y_p = u_1y_1 + u_2y_2 . \nonumber$

The Wronskian matrix is

$W = \begin{pmatrix} x^2 x^2 \ln x \\ 2x x + 2x \ln x \end{pmatrix}.\nonumber$

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

$w = x^3 + 2x^3 \ln x - 2x^3 \ln x = x^3.\nonumber$

So the inverse is

$W^{-1} = \dfrac {1}{x^3} \begin{pmatrix} x + 2x \ln x -x^2 \ln x \\ -2x x^2 \end{pmatrix}.\nonumber$

We have

$\begin{pmatrix} u'_1 \\ u'_2 \end{pmatrix} = \dfrac {1}{x^3} \begin{pmatrix} x + 2x \ln x -x^2 \ln x \\ -2x x^2 \end{pmatrix} \begin{pmatrix} 0 \\ \ln x \end{pmatrix}$

$= \dfrac {1}{x^3} \begin{pmatrix} -x^2 {\left ( \ln x \right )}^2 \\ x^2 \ln x \end{pmatrix} = \begin{pmatrix} -\dfrac{{(\ln x)}^2 }{x} \\ \dfrac {\ln x}{x} \end{pmatrix}\nonumber$

Integrating using u-substitution gives

$u_1 =\dfrac{-(\ln x)^3}{3}, \nonumber$

$u_2 = \dfrac{(\ln x)^2}{2}.\nonumber$

We have

$y_p = - \dfrac {1}{3} x^2 {\left ( \ln x \right )}^3 + \dfrac {1}{2} x^2 {\left ( \ln x \right )}^3 = \dfrac {1}{6}{\left ( \ln x \right )}^3.\nonumber$

Finally we get

$y = c_1 x^2 + c_2 x^2 \ln x + \dfrac{1}{6} (\ln x)^3. \nonumber$