Solving 3xy" - 4y' - xy = 0 with Frobenius

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I'm really getting stuck at this and I'm trying to read on it but it's confusing.

I need just a start-up for this equation which is to be solved with the method of frobenius.

3xy" - 4y' - xy = 0

Just need a start.

Any help is appreciated.

Thank you
 
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Start by substituting a power series in x for y. y=sum(a_k*x^k). Do the derivatives and equate coefficients of equal powers. Go!
 
done, then
 

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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