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I am having trouble solving the following differential equation: $$y''+(x^2-1)y'+2xy=0$$ To start off I employed the power-series method, so let $$y=\sum_{n=0}^{\infty}a_nx^n$$ $$y'=\sum_{n=1}^{\infty}na_nx^{n-1}$$ $$y''=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}$$ Next I plugged my power-series into my original DE: $$\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}+\sum_{n=1}^{\infty}na_nx^{n+1}-\sum_{n=1}^{\infty}na_nx^{n-1}+\sum_{n=0}^{\infty}2a_nx^{n+1}=0$$ When I matched the powers of $x$ I ended up with the following equation: $$\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n+\sum_{n=2}^{\infty}(n-1)a_{n-1}x^n-\sum_{n=0}^{\infty}(n+1)a_{n+1}x^n+\sum_{n=1}^{\infty}2a_{n-1}x^n=0$$ Next I tried to match the indeces, so naturally I took out the first 2 terms from $\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n$, the first 2 terms from $\sum_{n=0}^{\infty}(n+1)a_{n+1}x^n$, and the first term from $\sum_{n=1}^{\infty}2a_{n-1}x^n$.

At this point I end up with 2 recurrence relations: $$2a_2+6a_3x-a_0-2a_2x+2a_0x=0$$ $$(n+2)(n+1)a_{n+2}+(n-1)a_{n-1}-(n+1)a_{n+1}+2a_{n-1}=0$$

And here I am completely lost, normally what I would do is solve for one of the first $a_n$ terms but since I've got $x$'s in the first recurrence I can't really see how that would work. Perhaps there is a much simpler method that I can use? This is just practice for my ODE final exam.

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2 Answers 2

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Hint:

Notice that $(x^2-1)'=2x$, hence

$$y''+(x^2-1)y'+2xy=(y'+(x^2-1)y)'=0.$$

Par résolution of the first-order linear equation, $$y=e^{x-x^3/3}\left(C_2+C_1\int e^{x^3/3-x}dx\right).$$

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  • $\begingroup$ So that would yield only one solution though right? Would I use the power series to find the second solution? $\endgroup$
    – vovnsons
    Dec 9, 2019 at 20:51
  • $\begingroup$ @vovnsons: I gave you a double infinity of solutions. Isn't that enough ? $\endgroup$
    – user65203
    Dec 9, 2019 at 21:23
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Idea: $$(y''-y') +(x^2y'+2xy)=0$$ so $$(y'-y)' = -(x^2y)'\implies y'-y = -x^2y+c$$

Now solve that one. For $c=0$ you have $${dy\over y} = (1-x^2)dx$$

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