REQ Solution to differential equation

Kurret
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Find the solution to \frac{c}{y^2}+k=y''

I have not started with differential equations yet, so i hav eno idea if its possible to solve, but i stumbled upon it when investigating a physics idea i got. Thanks for any help!
 
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uhm, could anyone just leave a comment if its possible to solve or not?
 
well we can first sketch a solution for y''=k which is y=kx^2/2+Ax+B
but for: y''y^2=c I don't think there's an analytical answer.
 
Multiply by y' and you'll be able to integrate both terms of the new equation.
 
bigubau said:
Multiply by y' and you'll be able to integrate both terms of the new equation.
How?? Cant make it work...
 

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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