# Thread: diff equation problem help needed

1. ## diff equation problem help needed

(2y^2+5x)dx+3xydy=0 solve

2. Originally Posted by uga75

$\displaystyle (2y^2+5x)dx+3xydy=0$ solve
This equation can be made exact by finding an integrating factor

Let $\displaystyle \phi(x)$ be a differentable function of x

multiplying the equation by phi we get

$\displaystyle (2y^2+5x)\phi(x) dx+3xy \phi(x) dy=0$

Remember to be exact that if we have an eqaution of the form

$\displaystyle Mdx+Ndy=0$ it is exact if

$\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

$\displaystyle 4y\phi(x)=3x\phi(x)+3xy\frac{d\phi}{dx} \iff y\phi(x)=3xy\frac{d\phi}{dx} \iff \frac{1}{x}\cdot \frac{1}{x}dx=\frac{1}{\phi}d\phi$

Now solving for phi we get

$\displaystyle \frac{1}{3}\ln(x)=\ln(\phi) \iff \phi(x)=x^{1/3}$

So using this we get

$\displaystyle (2x^{1/3}y^2+5x^{4/3})dx+3x^{4/3}ydy=0$

This is now exact and can be solve using that method.

I hope this helps.

3. Originally Posted by uga75
$\displaystyle \frac{dy}{dx} + \frac{2}{3} \frac{y}{x} = - \frac{5}{3y}$