Using Undetermined Coefficients find the particular solution

Given that $xy^{\prime\prime}+(2+2x)y^{\prime}+2y=0$ has a solution $y_{1}(x)=x^{-1}$, find the general solution of the differential equation $xy^{\prime\prime}+(2+2x)y^{\prime}+2y=8e^{2x}$

I used Reduction of order to find the other solution.

Let $$y=x^{-1}v$$ Then $$y^{\prime}=-x^{-2}v+x^{-1}v^{\prime}$$ $$y^{\prime\prime}=2x^{-3}v-2x^{-2}v^{\prime}+x^{-1}v^{\prime\prime}$$

Substituting what we have, we get:

$$v^{\prime\prime}+2v^{\prime}=0$$

Let $v^{\prime}=u\;$ so,

$$u=e^{-2x}$$

And finally we get,

$$v = \dfrac{-1}{2}e^{-2x}$$

This means,

$$y = \dfrac{-1}{2}e^{-2x}x^{-1}$$

So we get complementary solution or general solution of homogenous DE in the form of

$$y_{c}=c_{1}x^{-1}-c_{2}\dfrac{1}{2}e^{-2x}x^{-1}$$

But we have to find the general solution of the non-homogenous DE. So I have used Undetermined Coefficients to find the coefficient of particular solution but I get zero terms, I can't make it done. Thanks in advance!

Solution 1:

Substituting $y_p = \frac{c_1(x)}{x}+\frac{c_2(x)}{2x}e^{-2x}$ into the complete ODE we obtain

$$ 2c_1'(x)+c_1''(x)-8e^{2x}-\frac 12e^{-2x}\left(2c_2'(x)-c_2''(x)\right)=0 $$

now as $c_1,c_2$ are independents, we choose those which obey

$$ \cases{ 2c_1'(x)+c_1''(x)-8e^{2x}=0\\ 2c_2'(x)-c_2''(x)=0 } $$

obtaining as solutions

$$ \cases{ c_1(x) = e^{2x}+\mu_1e^{-2x}+\mu_2\\ c_2(x) = \mu_3 e^{2x}+\mu_4 } $$

from those we choose a particular solution so we follow with $\mu_1=\mu_2=\mu_3=\mu_4 = 0$ or $c_1(x) = e^{2x},\ c_2(x)=0$ and finally we have

$$ y =\frac {c_1}{x}+\frac{c_2}{x}e^{-2x}+\frac 1x e^{2x} $$