# Seperable Intial-Value Problem

• Jan 30th 2011, 09:38 AM
JJ007
Seperable Intial-Value Problem
$\displaystyle (x^2+3y^2)dx + (2xy)dy = 0$
$\displaystyle y(2)=6$

so this homogeneous eqn. needs to be transformed into a separable eqn. right?
$\displaystyle y=vx , v=\frac{y}{x}$

$\displaystyle \frac{dy}{dx}=\frac{x^2+3y^2}{2xy}$ or is it $\displaystyle \frac{x^2+3y^2}{2x}dx - \frac{y}{1}dy$= How can I plug in $\displaystyle \frac{y}{x}$ here ?

$\displaystyle \frac {dy}{dx}= v + x\frac{dv}{dx}$

Not sure how to solve this one.
• Jan 30th 2011, 09:53 AM
TheEmptySet
Quote:

Originally Posted by JJ007
$\displaystyle (x^2+3y^2)dx + (2xy)dy = 0$
$\displaystyle y(2)=6$

so this homogeneous eqn. needs to be transformed into a separable eqn. right?
$\displaystyle y=vx , v=\frac{y}{x}$

$\displaystyle \frac{dy}{dx}=\frac{x^2+3y^2}{2xy}$= How can I plug in $\displaystyle \frac{y}{x}$ here ?

$\displaystyle \frac {dy}{dx}= v + x\frac{dv}{dx}$

Not sure how to solve this one.

Multiply the numerator and denominator by

$\displaystyle \frac{1}{x^2}$

This gives

$\displaystyle \displaystyle \frac{1+3\left( \frac{y}{x}\right)^2}{2\left( \frac{y}{x}\right)}$
• Feb 4th 2011, 05:00 AM
dwsmith
Quote:

Originally Posted by JJ007
$\displaystyle \frac {dy}{dx}= v + x\frac{dv}{dx}$

You should write the above as follows:

$\displaystyle y=vx\Rightarrow dy=xdv+vdx$

You would then plug this into dy of the DE.