1
$\begingroup$

The differential equation is $$\frac{d^2y(x)}{dx^2}+\frac{1}{x}\frac{dy(x)}{dx}-\frac{m^2}{x^2}y(x)=\frac{1}{1-x}$$ I'm given $y_1(x)=x^m$.

So, using $P(x) = \frac{1}{x}$ I found $W(x)=W_0e^{-\int{dxP(x)}}=\frac{W_0}{x}$.

I then used $W(x)$ and $y_1(x)$ to find $y_2(x)=y_1(x)\int{}dx\frac{W(X)}{y_1^2(x)}=-\frac{W_0}{m}+W_0cx^m$.

Does this all follow so far?

Now I want to find the particular solution $y_p(x)=-y_1(x)\int{dx\frac{y_2(x)d(x)}{a(x)W(x)}}+y_2(x)\int{dx\frac{y_1(x)d(x)}{a(x)W(x)}}$, where $d(x)=\frac{1}{1-x}$ and $a(x)=1$.

When I try to do this, it's extremely messy. I figured I messed up in one of the above steps. Can anyone confirm that or give me a clue as to how to solve this?

$\endgroup$
0
$\begingroup$

Hint:

$y_1=x^m$ then $$y_2=y_1\int\frac{e^{-\int p\,dx}}{y_1^2}=x^m\int\frac{e^{-\int\frac1x\,dx}}{x^{2m}}=x^m\frac{x^{-2m}}{-2m}=\frac{1}{-2mx^m}$$ the general solution is $$y_p=C_1x^m+C_2\frac{1}{-2mx^m}$$

$\endgroup$
  • $\begingroup$ That's the general solution, but I'm looking for the particular solution. The integral is giving me some hypergeometric function which doesn't seem like something I should be getting. $\endgroup$ – Spuds Feb 8 '17 at 19:03
  • $\begingroup$ This problem is which book selected.? $\endgroup$ – Nosrati Feb 8 '17 at 20:02
  • $\begingroup$ It's from Mathematical Physics by Kusse and Westwig. $\endgroup$ – Spuds Feb 8 '17 at 22:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.