Solve Power Series: (x^2)(y") + y = 0

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Homework Statement



solve using power series:
(x^2)(y")+y=0


Homework Equations





The Attempt at a Solution



after solving it i stopped at :

an[n^2-n+1]=0
 
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Your problems are coming from the fact that the differential equation is singular at 0. You can't expand it in a power series around x=0. You'll have to expand around another point if you want to do it that way. For example let u=x+1. Then the equation becomes (u-1)^2*y''+y=0. Now expand it as a series in u.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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