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So I was reading back over my course problem sheet for differential equations ( I'm studying for exams right now) and I came across this question:

Given that

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

For $|x|\leq1$

Find a particular solution of the equation

$$(1-x^2)y''-2xy'=\tfrac{2}{(5-4x)^{1/2}} -1$$

I've been wracking my brains trying to solve it but I can't figure out how. From my research I found that one way to solve a second order linear inhomogenous differential equation ( as this one is ) is to use the method of variation of parameters. However my teacher assured me that this could be solved using only material covered in the notes and this method is not in them.

Here's what we covered up to the point of the notes that corresponded to this question. He also said that we should be able to solve any question in the course without referring back to our first course in solving o.d.e.'s which thought us methods such as froebenius, characteristic equations, integrating factor method etc ....

  1) Linear independence of functions

     a) Criteria for checking linear independence

     b)Finding a second solution 


  2) Orthogonal function expansions (Sturm-Liouville theorem)


    a) Self adjoint form

    b) linear independence  

    c)Orthogonality 

    d)Completeness 

So my question is how can we use the concepts I've listed to solve this equation , or is it even possible to do so ? Even though he said it was it seems like there's just a few ingredients missing!

Note:just to be clear this isn't a homework question it's just from a problem sheet i'm studying for my exam in a few weeks.

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  • $\begingroup$ You can do it in an eigenfunction expansion by rewriting the homogenous ODE into Sturm-Liouville using an integration factor. $\endgroup$ – Story123 Dec 28 '18 at 0:38
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The $P_n(x)$'s are the Legendre polynomials, which satisfy $$ (1 - x^2)P_n''(x) - 2xP_n'(x) = -n(n+1)P_n(x).$$

Therefore, if $f(x)$ is the function $$ f(x) = \sum_{n = 0}^\infty a_n P_n(x),$$ then the solution to the differential equation $$ (1 - x^2)y'' - 2xy' = f(x)$$ must be $$ y=- \sum_{n=0}^\infty \frac{a_n}{n(n+1)}P_n(x).$$

Your question is a special case, with $a_n = 1 / 2^{n} - \delta_{n, 0}$.

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  • $\begingroup$ So all you have to do is take the LHS of your first equation then say : $-n(n+1)y=\sum a_n P_n(x) \Rightarrow y= - \sum_{n=0}^\infty \frac{a_n}{n(n+1)}P_n(x) $ ? That's a lot easier than I thought it was going to be ! Do you perhaps have any good links to literature regarding particular solutions to inhomogeneous versions of Legendre, Hermite, Laguerre etc .. It wasn't covered well in class at all and I'd like to do some extra study on it. By the way thank you so much for your post , I would have spent hours going down the wrong path had you not shown me how simple it was :) $\endgroup$ – excalibirr Dec 28 '18 at 1:04
  • $\begingroup$ @can'tcauchy Try plugging my solution into the equation, and verify that it satisfies the equation! You'll see that $$ \left[ (1 - x^2) \frac{d^2}{dx^2} - 2x \frac{d}{dx}\right] \left( - \frac{a_n }{n (n+1)} P_n (x)\right) = a_n P_n(x)$$ holds for each $n$. $\endgroup$ – Kenny Wong Dec 28 '18 at 1:06
  • $\begingroup$ @can'tcauchy Important and related revision exercise: Solve the equation for arbitrary $f(x)$. Hint: use the orthogonality properties of the Legendre polynomials to show that $f(x) = \sum_{n = 0}^\infty a_n P_n (x)$, where $a_n = \frac{2n + 1}{ 2} \int_{-1}^1f(x) P_n (x)$. (See here: en.wikipedia.org/wiki/…) Then use the same technique... $\endgroup$ – Kenny Wong Dec 28 '18 at 1:10
  • $\begingroup$ thank you this all helps a lot :) By chance I don't suppose you know of any links or even books that have any worked examples of questions like this do you ? I usually learn the underlying concepts best by seeing them applied ? $\endgroup$ – excalibirr Dec 28 '18 at 1:17
  • $\begingroup$ @can'tcauchy Perhaps Riley Hobson and Bence? $\endgroup$ – Kenny Wong Dec 28 '18 at 1:25

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