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.

  • 2
    $\begingroup$ It's $y = c_{1}e^{-2x}+c_{2}xe^{-2x}$ $\endgroup$
    – MtGlasser
    Mar 25 '20 at 2:24

The easiest way: $$y''+4y'+4y=\frac{e^{-2x}}{\sqrt{x^2 - 1}}$$ $$e^{2x}(y''+4y'+4y)=\frac{1}{\sqrt{x^2 - 1}}$$ $$(e^{2x}y)''=\frac{1}{\sqrt{x^2 - 1}}$$ Integrate both sides twice.

  • $\begingroup$ Im just unsure about something $\endgroup$ Mar 25 '20 at 4:10
  • $\begingroup$ You're welcomed. What is it ? @N_Mathematics_B $\endgroup$
    – MtGlasser
    Mar 25 '20 at 4:10
  • $\begingroup$ how did you go from line 2 to line 3, how did $$e^{2x}(y''+4y'+4y)$$ become $$(e^{2x}y)''$$ $\endgroup$ Mar 25 '20 at 4:11
  • $\begingroup$ If you can't go from step 2 to step 3 try to gro from step 3 to step 2....Differentiate lign 3 and see what you get..@N_Mathematics_B $\endgroup$
    – MtGlasser
    Mar 25 '20 at 4:12
  • $\begingroup$ here you can substitute $y=ze^{-2x}$ then the differential equation willl look simple....Thats another method...@N_Mathematics_B $\endgroup$
    – MtGlasser
    Mar 25 '20 at 4:16

$$y''-4y'+4y=\frac{e^{-2x}}{\sqrt{x^2-1}}=f(x)$$ $r=-2$ 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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