Second Order Differential Equation problem with initial value. Can't integrate further?

Question

$y$ double prime $- 3y^2 = 0$

$y$ prime $(0) = 4$

$ y (0) = 2$

So this question popped up in my textbook that has me stuck.

My attempt,

Let y double prime = $v (dv/dy)$

So it becomes,

$v (dv/dy) - 3y^2 = 0$

$v * dv = 3y^2 * dy$

Integrate both sides,

$(1/2)v^2 = y^3 + c$

$v = (2y^3 + 2c)^{1/2}$

Since $v = y$ prime and $y$ prime $(0) = 4$

$4 = ( 0 + 2c )^{1/2}$

$c = 8$

$dy/dx = (2y^3 + 16) ^{1/2}$

$dy (2y^3 + 16)^{-1/2} = dx$

Now...I can't integrate the left side for the life of me.....

Any help is appreciated!

Answer 1

At this line $4 = ( 0 + 2c )^{1/2}$ It seems you made a mistake with the constant c $y(0)^3=8$ not 0

$$y''-3y^2=0$$ substitute $p=\frac {dy}{dx} \implies y''=p\frac {dp}{dy}=pp'$ $$pp'-3y^2=0$$ $$pdp=3y^2dy$$ After integration $$\frac {p^2}2=y^3+K$$ $$y'(0)=4 \, ,\, y(0)=2 \implies K=0$$ $$ {y'^2}=2y^3$$