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} $.