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Use power series to solve the initial-value problem: $$y''-2xy'-2y=0; y(0)=1, y'(0)=0$$

How to we solve this in terms of summation sign?

I got this :

$a_{n+2}=2\frac{a_n}{n+2}$

$y(x)=c_0(1-x^2-\frac12x^4)+c_1(x+\frac23x^3+\frac4{15}x^5)$

I dont know what will it be in terms of odd and even summations.

Paul Sinclair
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jelli
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  • Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. – saulspatz Nov 28 '20 at 04:45
  • Hint: the answer is $\exp(x^2)$. – Parcly Taxel Nov 28 '20 at 04:56
  • I am getting (-1)^n/n! for even numbers.I dont think I am getting the right series – jelli Nov 28 '20 at 05:30

1 Answers1

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$$a_{n+2}=2\frac{a_n}{n+2};\;a_0=1;\;a_1=0$$ $$\begin{cases} a_{2n+1}=0\\ a_{2n}=\frac{1}{n!}\\ \end{cases} $$ $$y=\sum _{n=0}^{\infty } \frac{x^{2 n}}{n!}=e^{x^2}$$

Raffaele
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