# Math Help - differential equations-help!!!!!!!

1. ## differential equations-help!!!!!!!

2. Hello, mbempeni!

The first one is homegeneous . . .

$\frac{dy}{dx} \;=\;\frac{x^3 + 2x^2y - y^3}{x^3-xy^2}$
Divide top and bottom of the fraction by $x^3\!:\quad\frac{dy}{dx} \;=\;\frac{1 - 2\left(\dfrac{y}{x}\right) - \left(\dfrac{y}{x}\right)^3}{1 - \left(\dfrac{y}{x}\right)^2}$

Let $v \,=\,\frac{y}{x}\quad\Rightarrow\quad y \,=\,vx\quad\Rightarrow\quad \frac{dy}{dx} \:=\:v + x\frac{dv}{dx}$

Substitute: . $v + x\frac{dv}{dx} \;=\;\frac{1-2v-v^3}{1-v^2}$

Then: . $x\frac{dv}{dx} \;=\;\frac{1-2v-v^3}{1-v^2} - v \quad\Rightarrow\quad x\frac{dv}{dx}\;=\;\frac{1-3v}{1-v^2}$

Separate variables: . $\frac{v^2-1}{3v-1}\,dv \;=\;\frac{dx}{x}$

Can you finish it now?

3. Originally Posted by Soroban
Hello, mbempeni!

The first one is homegeneous . . .

Divide top and bottom of the fraction by $x^3\!:\quad\frac{dy}{dx} \;=\;\frac{1 - 2\left(\dfrac{y}{x}\right) - \left(\dfrac{y}{x}\right)^3}{1 - \left(\dfrac{y}{x}\right)^2}$

Let $v \,=\,\frac{y}{x}\quad\Rightarrow\quad y \,=\,vx\quad\Rightarrow\quad \frac{dy}{dx} \:=\:v + x\frac{dv}{dx}$

Substitute: . $v + x\frac{dv}{dx} \;=\;\frac{1-2v-v^3}{1-v^2}$

Then: . $x\frac{dv}{dx} \;=\;\frac{1-2v-v^3}{1-v^2} - v \quad\Rightarrow\quad x\frac{dv}{dx}\;=\;\frac{1-3v}{1-v^2}$

Separate variables: . $\frac{v^2-1}{3v-1}\,dv \;=\;\frac{dx}{x}$

Can you finish it now?
Can you finish it?It much more difficult than I thought.Please..

4. Originally Posted by mbempeni
Can you finish it?It much more difficult than I thought.Please..
Surely you can integrate the right hand side.

As for the left hand side, note that $\frac{v^2 - 1}{3v - 1} = \frac{v}{3} + \frac{1}{9} - \frac{8}{9} \left( \frac{1}{3v - 1}\right)$.

If you're studying differential equations then it's expected that you can apply the various techniques of integration that will be required and which you have no doubt been taught. It might be wise to go back and revise them.