# Differential Equations: Separable Equations

1. Jan 23, 2014

### KallKoll

1. The problem statement, all variables and given/known data
Solve the equation

dy/dx = x/(y^2√(1+x))

2. Relevant equations

3. The attempt at a solution
I separated them:
y^2 dy = dx/√(1+x)

I then integrated the dy side, I got (1/3)y^3 + C. I am stuck at integrating the dx side. Thanks in advance!

2. Jan 23, 2014

### Curious3141

Shouldn't that be $\int y^2dy = \int \frac{x}{\sqrt{1+x}}dx$?

For the RHS, try $u^2 = 1 + x$.

3. Jan 23, 2014

### SammyS

Staff Emeritus

You dropped an x from the numerator on the right side.

4. Jan 23, 2014

### KallKoll

Yes that should be, thank you. How does the u^2 work?

5. Jan 23, 2014

### Curious3141

Did you try it? Post what you have. Latex formatted please.

6. Jan 24, 2014

### KallKoll

Well, I set $\sqrt{1+x}$ = u. Then I set u2 = 1 + x. But I don't understand how to get du from that or how that helps with getting rid of the x on top.

7. Jan 24, 2014

### pasmith

$$\frac{x}{\sqrt{1 +x}} = \frac{u^2 - 1}{u} = u - \frac1u \\$$

8. Jan 24, 2014

### HallsofIvy

Staff Emeritus
From $u^2= 1+ x$, $x= u^2- 1$ and $2u du= dx$ so
$$\int \frac{xdx}{\sqrt{1+ x}}= \int \frac{(u^2- 1)(2udu)}{u}= 2\int (u^2- 1) du$$

9. Jan 24, 2014

### KallKoll

Thank you so much!