# Help with Homogeneous/Seperable equation

• December 4th 2009, 02:00 PM
cYn
Help with Homogeneous/Seperable equation
OK so here is my problem:

$\frac{dy}{dx} = \frac{(x+y)^2}{2x}$

My steps:

$y = vx$
$y' = v + v'x$
$y' = \frac{(x+y)^2}{2x}$
$v + v'x = \frac{(x+y)^2}{2x}$
$v + \frac{dv}{dx}x = \frac{(x+y)^2}{2x}$
$v + \frac{dv}{dx}x = \frac{(x+vx)^2}{2x}$
$v + \frac{dv}{dx}x = \frac{x^2 + 2vx^2 +v^2x^2}{2x}$
$v + \frac{dv}{dx}x = \frac{x^2}{2x} + \frac{2vx^2}{2x} + \frac{v^2x^2}{2x}$
$v + \frac{dv}{dx}x = \frac{x}{2} + vx + \frac{v^2x}{2}$

If I divide both sides by x I get:

$\frac{v}{x} + \frac{dv}{dx} = \frac{1}{2} + v + \frac{v^2}{2}$
$\frac{v}{x} + dv = (\frac{1}{2} + v + \frac{v^2}{2})dx$

This is the part that I'm having trouble with. I haven't done homogeneous/seperable in a long time. Can someone help me with the next step? And possibly a solution?
• December 4th 2009, 08:51 PM
TheEmptySet
Quote:

Originally Posted by cYn
OK so here is my problem:

$\frac{dy}{dx} = \frac{(x+y)^2}{2x}$

My steps:

$y = vx$
$y' = v + v'x$
$y' = \frac{(x+y)^2}{2x}$
$v + v'x = \frac{(x+y)^2}{2x}$
$v + \frac{dv}{dx}x = \frac{(x+y)^2}{2x}$
$v + \frac{dv}{dx}x = \frac{(x+vx)^2}{2x}$
$v + \frac{dv}{dx}x = \frac{x^2 + 2vx^2 +v^2x^2}{2x}$
$v + \frac{dv}{dx}x = \frac{x^2}{2x} + \frac{2vx^2}{2x} + \frac{v^2x^2}{2x}$
$v + \frac{dv}{dx}x = \frac{x}{2} + vx + \frac{v^2x}{2}$

If I divide both sides by x I get:

$\frac{v}{x} + \frac{dv}{dx} = \frac{1}{2} + v + \frac{v^2}{2}$
$\frac{v}{x} + dv = (\frac{1}{2} + v + \frac{v^2}{2})dx$

This is the part that I'm having trouble with. I haven't done homogeneous/seperable in a long time. Can someone help me with the next step? And possibly a solution?

If you expand out the right hand side you will get an equation of the form.

$\frac{dy}{dx}=P(x)+Q(x)y+R(x)y^2$

This is know as Ricatti's equation see here

Riccati equation - Wikipedia, the free encyclopedia

The page has a method to solve it.

Good luck.