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what is the method for solving a differential equation when a summation is involved from the start ?

ex. what method is required to find a particular solution to

$$(1-x^2)y''-2xy'=\sum_{n=1}^{\infty}\tfrac{P_n(x)}{2^n}$$

I should point out also that

$\tfrac{2}{(5-4x)^{1/2}}=\sum_{n=0}^{\infty}\tfrac{P_n(x)}{2^n}$.

Edit: $|x| \leq 1$

My attempt :

$\tfrac{2}{(5-4x)^{1/2}}=\sum_{n=0}^{\infty}\tfrac{P_n(x)}{2^n} \Rightarrow \sum_{n=0}^{\infty}\tfrac{2^{n+1}}{(5-4x)^{1/2}}=\sum_{n=0}^{\infty}P_n(x)$. ( I don't know if you can do this though.....)

then we assume that $y=\sum_{n=0}^{\infty}a_nP_n(x)=\sum \tfrac{a_n2^{n+1}}{(5-4x)^{1/2}}$

$y'=\sum \tfrac{a_n2^{n+2}}{(5-4x)^{3/2}}$

$y''=\sum \tfrac{3a_n2^{n+3}}{(5-4x)^{5/2}}$

Is this anyway right ?

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  • $\begingroup$ Is the left side $((1-x^2)y')'$? $\endgroup$ – Nosrati Oct 30 '18 at 13:00
  • $\begingroup$ @Nosrati I think that is a way to express it yes . Its the legendre equation anyway . $\endgroup$ – excalibirr Oct 30 '18 at 13:01
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It doesn't matter what function you have on the right hand side. Let's call it:

$$f(x)=\sum_{n=1}^{\infty}\frac{P_n(x)}{2^n}$$

As @Nosrati correctly noted, you have:

$$(1-x^2)y''-2xy'=((1-x^2)y')'$$

Which immediately allows us to integrate both sides:

$$(1-x^2)y'=\int_a^x f(t) dt$$

Where $a$ is an arbitrary constant, which depends on the intitial conditions. I have written the integral this way to keep track of the different integration variables we are going to have.

Now we assume $x \neq \pm 1$ and find:

$$y(x)=\int_b^x \frac{1}{1-u^2} \int_a^u f(t) dt ~du$$

Where the second constant $b$ also depends on the initial conditions.

Now whatever the definition of $f(x)$ (a series or a closed form), you just need to evaluate the integrals and apply the initial conditions correctly.

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  • $\begingroup$ I should have pointed out in my main body that we are also told that $|x| \leq 1$. how does this change the method you suggested. I'm sorry, I thought i had included that information originally $\endgroup$ – excalibirr Oct 30 '18 at 21:02
  • $\begingroup$ @exodius, deal with the case $x^2=1$ separately. You can consider this solution valid for $|x|<1$ $\endgroup$ – Yuriy S Oct 30 '18 at 21:06
  • $\begingroup$ your answer springboarded the following work. So we have that $((1-x^2)y')'=\sum_{n=1}\tfrac{P_n(x)}{2^n}$. But by rodriguezes formula we have that $P_n(x)=\tfrac{1}{2^nn!}\tfrac{d^n}{dx^n}(x^2-1)^n$. which means that $((1-x^2)y')'=\sum_{n=1}\tfrac{1}{4^nn!}\tfrac{d^n}{dx^n}(x^2-1)^n$ Now we can integrate both sides with respect to x $$(1-x^2)y'=\int^1_{-1}\sum_{n=1}\tfrac{1}{4^nn!}\tfrac{d^n}{dx^n}(x^2-1)^n=\sum_{n=1}\tfrac{1}{4^nn!}[\tfrac{d^{n-1}}{dx^{n-1}}(x^2-1)^n]^1_{-1}$$ $\endgroup$ – excalibirr Oct 30 '18 at 22:04
  • $\begingroup$ But obviously the value on the right will be zero for all n. what did you mean by seperating the cases where |x|<1 and $x^2=1$. also in your example how do you know what values to sub in for x,b,a,u . I think that perhaps where i have it here that u=-a must be incorrect as it makes the whole thing equal zero $\endgroup$ – excalibirr Oct 30 '18 at 22:04
  • $\begingroup$ @exodius, you are supposed to integrate w.r.t. $t$ and then w.r.t. $u$. That way you get a function of a single variable $x$, as it should be. $a,b$ are constants, determined by the initial conditions. You can treat them as arbitrary parameters, and then after getting the general expression for $y(x)$ you compare it to your initial conditions and solve for $a,b$. $\endgroup$ – Yuriy S Oct 30 '18 at 22:08

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