## RolyPoly 2 years ago $\frac{dy}{dx} = \frac{x}{y^2\sqrt{1+x}}$

1. RolyPoly

$\frac{dy}{dx} = \frac{x}{y^2\sqrt{1+x}}$$y^2dy = \frac{x}{\sqrt{1+x}}dx$$\frac{1}{3}y^3= \int \frac{u-1}{u}du = \frac{2}{3} u^{\frac{3}{2}}- 2u^{\frac{1}{2}}+C$$y = 2(1+x)^{\frac{3}{2}}-6(1+x)^{\frac{1}{2}}+C)^{\frac{1}{3}}$

2. hartnn
3. RolyPoly

Did I do something wrong again?

4. hartnn

i couldn't understand clearly what u did, but it can be simplified to 2/3*(x-2)*sqrt(x+1)

5. RolyPoly

Which part you didn't understand?

6. hartnn

u=?

7. RolyPoly

1+x

8. hartnn

then denominator is sqrt u

9. RolyPoly

OMG!I typed something wrong!! $\frac{1}{3}y^3= \int \frac{u-1}{\sqrt u}du = \frac{2}{3} u^{\frac{3}{2}}- 2u^{\frac{1}{2}}+C$

10. RolyPoly

No wonder why you couldn't understand.. That was my mistake :( Is that clear now?

11. hartnn

12. hartnn

if needed.

13. RolyPoly

That's how the answer in my book looks like :S But thanks!

14. hartnn

ohh.. then no need. you were correct all along. welcome ^_^