3.5: 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 \(y_1\) 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

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

Finally we get

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

Contributors and Attributions

• Larry Green (Lake Tahoe Community College)

• Integrated by Justin Marshall.