Help Solving ODE: y'' - 4k^2 y^2 = 0

[eckiller]

I need help completing this problem:

y'' - 4k^2 y^2 = 0, y(0) = 2, y'(0) = 4k, k > 0 and fixed constant

Let v = y'

dv/dx = dv/dy * dy/dx = dv/dy * v

dv/dy * v - 3k^2 y^2 = 0

Integral( v dv ) = Integral( 3k^2 y^2 )

v^2 = 2k^2 y^3 + C0

v(0) = 4k ==> C0 = 16k^2

dy/dx = sqrt( 2k^2 y^3 + 16k^2 )

I think this is separable, but I can't do the integral, so I think I messed up somewhere.

 

[daveb]

Your equation for v(0) is wrong, since v(0)=4k ==> 16k^2 = 2k^2*y(0)^3 + C0 = 2k^2 * 8 * C0 ==> C0 = 0. This makes the separable equation easier.