The method of variation of parameter (II)

Problem:

$y'' - 3y' + 2y = \cos x$

Solution:

The solution to the homogeneous equation $w'' - 3w' + 2w = 0$ is obviously $w_1 = e^x$ and $w_2 = e^{2x}$.

Next, we solve the equation pair:

$v_1' e^x + v_2' e^{2x} = 0$.
$v_1' e^x + 2 v_2' e^{2x} = \cos x$.

The second equation subtracts the first equation gives:

$v_2' e^{2x} = \cos x$

Therefore $v_2' = \cos x e^{-2x}$.

This time I do the integral myself using Euler's relation!

Consider the integral $\int{e^{ix}e^{-2x}dx} = \int{(\cos x + i\sin x)e^{-2x}dx}$. Therefore we can equate the real part to get the result, and the first integral is particularly easy.

$\int{e^{ix}e^{-2x}dx} = \frac{1}{-2 + i}e^{(-2 + i)x}$.

Next we find the real part of it by rationalizing and Euler's relation.

$\begin{eqnarray*} & & \frac{1}{-2 + i}e^{(-2 + i)x} \\ &=& \frac{(-2 - i)}{(-2 + i)(-2 - i)}e^{(-2 + i)x} \\ &=& \frac{(-2 - i)}{5}e^{(-2 + i)x} \\ &=& \frac{(-2 - i)}{5}e^{-2x}e^{ix} \\ &=& \frac{(-2 - i)}{5}e^{-2x}(\cos x + i\sin x) \\ \end{eqnarray*}$

Therefore, we have

$\begin{eqnarray*} & & v_2(x) \\ &=& \int \cos x e^{-2x} dx \\ &=& e^{-2x}(\frac{-2}{5}\cos x + \frac{1}{5}\sin x) \end{eqnarray*}$

Similarly, we solve $v_1'(x)$ by subtracting the second equation by twice the first one, that gives $v_1'(x) = -\cos(x)e^{-2x}$. The integral is very similar and we get $\begin{eqnarray*} & & v_1(x) \\ &=& \int -\cos x e^{-x} dx \\ &=& e^{-x}(\frac{1}{2}\cos x - \frac{1}{2}\sin x) \end{eqnarray*}$

Therefore the answer is $(\frac{-2}{5} + \frac{1}{2})\cos x + (\frac{1}{5} - \frac{1}{2})\sin x + Ae^x + Be^{2x} = \frac{1}{10}\cos x - \frac{3}{10}\sin x + Ae^x + Be^{2x}$.