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I have been given the differential equation ${d^2\over dx^2}y-x^2{d^2\over dx}y-xy=0$

and I have been attempting to solve with the power series method. I have worked tried working a few different ways and have gotten different answers but I feel that this way was the most sound. If someone could check my work and see if I am going about it the right way or if not hint me onto the right path that would be much appreciated. Thank you.

I first started with a solution of the form $$y=\sum_{n=0}^\infty A_nx^n$$ and then differentiated twice to get $y'=\sum_{n=1}^\infty nA_nx^{n-1}$ and $y''=\sum_{n=2}^\infty n(n-1)A_nx^{n-2}$

and after substituting back in $$\sum_{n=2}^\infty n(n-1)A_nx^{n-2}-x^2\sum_{n=1}^\infty nA_nx^{n-1}-x\sum_{n=0}^\infty A_nx^n=0$$ $$\sum_{n=2}^\infty n(n-1)A_nx^{n-2}-\sum_{n=1}^\infty nA_nx^{n+1}-\sum_{n=0}^\infty A_nx^{n+1}=0$$ And now the left sum I shift up by two and the middle sum I can start at $n=0$ and have it not change the sum. $$\sum_{n=0}^\infty (n+2)(n+1)A_{n+2}x^n-\sum_{n=0}^\infty nA_nx^{n+1}-\sum_{n=0}^\infty A_nx^{n+1}=0$$ Now I can shift both the middle and right sum down by one. $$\sum_{n=0}^\infty (n+2)(n+1)A_{n+2}x^n-\sum_{n=1}^\infty (n-1)A_{n-1}x^n-\sum_{n=1}^\infty A_{n-1}x^n=0$$ So now that I have all the sums being multiplied by $x^n$ I can pull a term out of the left sum so they all start at $n=1$ and then bring them all under one sum and factor out the $x^n$ $$2A_2+\sum_{n=1}^\infty [(n+2)(n+1)A_{n+2}-(n-1)A_{n-1}-A_{n-1}]x^n=0$$ now I find $A_2=0$ and a recurrence relation $$A_{n+2}={n\over (n+2)(n+1)}A_{n-1}$$

and the first four terms are $$A_0$$ $$A_1$$ $$A_2=0$$ $$A_3={1\over 6}A_0$$

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    $\begingroup$ As I saw it's correct and then ! $\endgroup$ – Nosrati Nov 17 '17 at 21:43
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    $\begingroup$ $$y=A_0\left(1+\dfrac16x^3+\dfrac{1}{45}x^6+\cdots\right) +A_1\left(x+\dfrac16x^4+\dfrac{5}{252}x^7+\cdots\right)$$ $\endgroup$ – Nosrati Nov 17 '17 at 21:54
  • $\begingroup$ Yes, looks fine to me. $\endgroup$ – Malcolm Nov 17 '17 at 22:00
  • $\begingroup$ Thanks for the help guys! $\endgroup$ – jimmyoung Nov 20 '17 at 2:37

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