Variation of Parameters

Problem

Use the variation of parameters method to solve the ordinary differential equation

$$y'' - y' - 2y = 4{e}^{-t}.$$

Solution

To solve for the complementary solution, determine the roots of the characteristic equation

$$D^2 - D - 2 = 0.$$

This characteristic equation can be factored $(D-2)(D+1) = 0$; thus, the roots are $D=2$ and $D=-1$.

Hence the complementary solution is $y_c (t) = A {e}^{2t} + B {e}^{-t}$. Here, $A, B$ are arbitrary constants.

The next step is to calculate the Wronskian determinant.

Let $y_1 = {e}^{2t}$ and $y_2 = {e}^{-t}$. Taking the derivatives of each function yields $y_1 = 2{e}^{2t}$ and $y_2 = -{e}^{-t}$.

$$W = \begin{vmatrix} y_1 & y_2 \\ y'_1 & y'_2 \end{vmatrix}$$

$$W = \begin{vmatrix} {e}^{2t} & {e}^{-t} \\ 2{e}^{2t} & -{e}^{-t} \end{vmatrix}$$

$$W = ({e}^{2t})(-{e}^{-t}) - ({e}^{-t})(2{e}^{2t}) = -3 e^t$$

The formula for calculating the particular solution is

$$y_p (t) = -y_1 \left(\int \frac{y_2 g(t)}{W} dt \right) + y_2 \left(\int \frac{y_1 g(t)}{W} dt \right)$$

$$\begin{align} y_p (t) &= -{e}^{2t} \left(\int \frac{{e}^{-t} 4{e}^{-t}}{-3 e^t} \,dt \right) + {e}^{-t} \left(\int \frac{{e}^{2t} 4{e}^{-t}}{-3 e^t} \,dt \right) \\[5 pt] &= \frac{4}{3}{e}^{2t} \left(\int {e}^{-3t} \,dt \right) - \frac{4}{3}{e}^{-t} \left(\int 1\,dt \right) \\[5 pt] &= \frac{-4}{9} {e}^{-t} - \frac{4}{3} t{e}^{-t} \end{align}$$

There is a new term $- \frac{4}{3} t{e}^{-t}$appearing! Add this to the complementary solution to get

$$y_c (t) = A {e}^{2t} + B {e}^{-t} - \frac{4}{3} t{e}^{-t}.$$