Given (y')^2 *y'' -8y=0 and ICs y(0)=1 y'(0)= -2(adsbygoogle = window.adsbygoogle || []).push({});

I used substitution dy/dx = v and d^2y/dx^2 = vdv/dy and got to:

v^2*v*v' = 8y

V^3 dv/dy = 8y

v^4/4 =4y + c

V^4 =4y + c

And IC y'(0) = v(0) =-2

So c = 16

v = 2(y^2+1)^(1/4)

dy/dx = 2(y^2+1)^(1/4)

(y^2+1)^(-1/4)dy = 2dx

But here I got stuck. :( did I do the steps wrong? I then tried trig substitution and got:

Sec(u)^3/2 du = 2dx

But couldn't integrate either. This is a DE class so I didn't think it should be like that and maybe I made a mistake. Any help would be greatly appreciated. :]

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# Nonlinear ODE

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