# Second Order Nonlinear ODE

1. Sep 25, 2012

### josephsuk

1. y''y^4 = 8

I tried almost every method I know, including laplace transforms, variation of parameters, reductin of order, v=y' substitution

2. Sep 25, 2012

### jackmell

First, you realize that y'' can be written as d(y') and y' as dy right? So what happens if you multiply both sides of the equation by y'?

Well, you get:

$$y'd(y')y^4=8dy$$

Now what?

Last edited: Sep 25, 2012
3. Sep 25, 2012

### josephsuk

It appears to be logical to use separation of variables to get y'd(y')=8y^-4dy
and then integrate to get the solution?

Last edited: Sep 25, 2012
4. Sep 25, 2012

### jackmell

Forgot the other part. I get:

$$y'd(y')=8y^{-4} dy$$

5. Sep 25, 2012

### josephsuk

I am not exactly what follows in the next steps.

Would integrating y' d(y') give me ((y')^2)/2 or something else?

If it does give me that answer, then I get y to be $$y=\frac{5}{2}x\sqrt[5]{\frac{-16}{3}}$$

6. Sep 25, 2012

### LCKurtz

Look at your given DE and see if you think a constant times x can be a solution.

7. Sep 25, 2012

### jackmell

No way dude. you got:

$$\int y' d(y')=\int 8 y^{-4} dy$$
and that's
$$\frac{(y')^2}{2}=-8/3 y^{-3}+c$$
Ok, can you now separate variables (take square root first), and then post what the next integral expression would be? Can't integrate it directly (not easily) but just the expression of what has to be integrated is good enough for now.