# Nonlinear ODE

1. Feb 23, 2012

### thepatient

Given (y')^2 *y'' -8y=0 and ICs y(0)=1 y'(0)= -2

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. :]

2. Feb 23, 2012

### HallsofIvy

Staff Emeritus
The integral of 8y is NOT 4y+ C.

How did you get this? If you multiply both sides of v^4/4= 4y+ C (which was wrong) by 4 you would get y^4= 16y+ 4c (which you could write as y^4= 16y+ C' since 4 times an unknown constant is still an unknown constant).

3. Feb 23, 2012

### thepatient

Oops yes, I had before putting in first IC

V^4 = 16y^2 + C.

I jumped a bit a head of myself while typing this in. XD I just had this on an exam a few hours ago.

Last edited: Feb 23, 2012
4. Feb 23, 2012

### thepatient

From:
V^3 dv/dy = 8y, I used method of separable variables.

V^3 dv = 8ydy then integrate both sides:

V^4/4 = 8y^2/2 + c

5. Feb 23, 2012

### thepatient

I accidentally left the squared out on that part. XD it's not easy typing on phone.