TITLE: Second Order Linear ODE (general solution)
QUESTION [2 upvotes]: Find the general solution to the ODE shown below:
$$y''+4y'+4y=\frac{e^{-2x}}{\sqrt{x^2 - 1}}$$
Below is all of my progress so far:
Clearly, this is a second-order linear, non-homogenous ODE. So, the general solution can be written as:
$$y = y_h+y_p$$
So, yh:
$$y''+4y'+4y = 0$$
$$r^2+4r+4 = 0$$
$$r^2+2r+2r+4 = 0$$
$$r(r+2)+2(r+2) = 0$$
$$∴r=-2$$
So, this means that:
$$y = c_{1}e^{-2x}+c_{2}e^{-2x}$$
Now we must consider yp:
We can use the method of undetermined coefficients.
This is where I get stuck
I have no idea how to use the method of undetermined coefficients to find yp. Any help with this question would be much appreciated since I've been stuck on it for a long time. Thanks in advance.

REPLY [1 votes]: $$y''-4y'+4y=\frac{e^{-2x}}{\sqrt{x^2-1}}=f(x)$$
$r=-2$ is a double root so the general soluyions are $y_1=C_1 e^{-2x}, y_2=C_2 x e^{-2x}$. The particilar solution $y_p$ will be found by variation of $C_1$ and $C_2$ w.r.t. $x$.
Let $w(x)=[y_1 y'_2-y'_1 y_2]=e^{-2x}$.
Then $$C_1(x)=-\int \frac{y_2f(x) dx }{w(x)}+D_1= -\int \frac{xe^{-2x} e^{-2x} dx}{\sqrt{x^2-1}~ e^{-4x}}+D_1$$ $$\implies C_1(x)=-\int \frac{xdx}{\sqrt{x^2-1}}+D_1.$$
$$\implies C_1(x)=- \sqrt{x^2-1}+D_1.$$
Next, $$C_2(x)=\int \frac{y_1 f(x) dx}{w(x)}+D_2=\int\frac{e^{-2x} e^{-2x} dx}{e^{-4x}\sqrt{x^2-1}}+D_2.$$
$$\implies C_2(x)=\int \frac{dx}{\sqrt{x^2-1}}+D_2=\ln[x+\sqrt{x^2-1}+D_2$$
The total solution is  $$y=C_1(x) e^{-2x}+ C_2(x) x e^{-2x}$$
$$y=-\sqrt{x^2-1}~ e^{-2x}+ x\ln(x+\sqrt{x^2-1})e^{-2x}+D_1 e^{2x}+
D_2 xe^{-2x}$$
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