# Thread: [SOLVED] Solve Separable ODE

1. ## [SOLVED] Solve Separable ODE

Im stuck on this question, been trying to figure out on the train, just ended up with crossed off working out.

dy/dx = y^2 x

Ans: y = 2/C-x^2

I end up separating it to

| dy/y^2 = | xdx

| 1/y^2 dy = | xdx

|1/y^2 dy = 1/2 x^2 + C

but i get stuck with trying to integrate y, im missing something, do i have to use the substitution rule?

help appreciated

2. Let's just step through it. Okey-doke.

$\frac{dy}{dx}=xy^{2}$

Separate:

$\frac{1}{y^{2}}dy=xdx$

Integrate:

$\int\frac{1}{y^{2}}dy=\int xdx$

$\frac{-1}{y}=\frac{1}{2}x^{2}+C$

$y=\frac{-2}{x^{2}+2C}$

3. Originally Posted by galactus
Let's just step through it. Okey-doke.

$\frac{dy}{dx}=xy^{2}$

Separate:

$\frac{1}{y^{2}}dy=xdx$

Integrate:

$\int\frac{1}{y^{2}}dy=\int xdx$

$\frac{-1}{y}=\frac{1}{2}x^{2}+C$

$y=\frac{-2}{x^{2}+2C}$
Using alternative notation...

$\frac{dy}{dx} = xy^2$

$y^{-2}\,\frac{dy}{dx} = x$

$\int{y^{-2}\,\frac{dy}{dx}\,dx} = \int{x\,dx}$

$\int{y^{-2}\,dy} = \int{x\,dx}$

I'm sure you can go from here...

4. Originally Posted by Prove It
Using alternative notation...

$\frac{dy}{dx} = xy^2$

$y^{-2}\,\frac{dy}{dx} = x$

$\int{y^{-2}\,\frac{dy}{dx}\,dx} = \int{x\,dx}$

$\int{y^{-2}\,dy} = \int{x\,dx}$

I'm sure you can go from here...
Yep sure can guys, thanks alot