MHB How to Find a General Solution Using Variation of Parameters?

[ineedhelpnow]

Use the variation of parameters method to find a general solution of $x^{2}y''+xy'-9y=48x^{5}$

$m^{2}-9=0$

$(m+3)(m-3)=0$

$m=3,-3$

$y_{h}=c_{1}x^{-3}+c_{2}x^{3}$

$W=6/x$ Don't really know how to do wronskian with latex so i didnt include the steps. But i need help with the rest of this. i know $y_{p}=-y_{1}\int \frac{y_{2}r}{W}+y_{2}\int \frac{y_{1}r}{W}$

What's r?

 

[MarkFL]

Just so I can get up to speed:

We have a Cauchy-Euler equation, so we can try a solution of the form:

$$y(x)=x^r$$

Hence:

$$y'(x)=rx^{r-1}$$

$$y''(x)=r(r-1)x^{r-2}$$

And by substitution, we have:

$$x^2\left(r(r-1)x^{r-2}\right)+x\left(rx^{r-1}\right)-9\left(x^r\right)=0$$

$$r(r-1)x^r+rx^r-9x^r=0$$

$$r(r-1)+r-9=0$$

$$r^2-9=(r+3)(r-3)=0$$

And so, yes...the homogeneous solution is:

$$y_h(x)=c_1x^{-3}+c_2x^3$$

Okay, now what I would do here is assume the particular solution has the form:

$$y_p(x)=v_1(x)x^{-3}+v_2(x)x^3$$

And then solve the system:

$$x^{-3}v_1'+x^3v_2'=0$$

$$-3x^{-4}v_1'+3x^2v_2'=48x^3$$

Now, from the first equation, we find:

$$v_1'=-x^6v_2'$$

And substituting into the second equation, we obtain:

$$-3x^{-4}\left(-x^6v_2'\right)+3x^2v_2'=48x^3$$

$$6x^2v_2'=48x^3$$

$$v_2'=8x\implies v_2(x)=4x^2$$

Hence:

$$v_1'=-x^6\left(8x\right)=-8x^7\implies v_1(x)=-x^{8}$$

And so our particular solution is:

$$y_p(x)=\left(-x^{8}\right)x^{-3}+\left(4x^2\right)x^3=3x^5$$

And so the general solution is:

$$y(x)=y_h(x)+y_p(x)=c_1x^{-3}+c_2x^3+3x^5$$

 

[ineedhelpnow]

I just figured it out :D thanks anyway! your steps are way more organized than mine.