# Homogeneous equation

• Feb 1st 2010, 01:05 PM
steph3824
Homogeneous equation
Show that the given equation is homogeneous and solve the differential equation.

$dy/dx=x^2+3y^2/2xy
$

To show that it is homogeneous, I have $dy/dx=1+3(y/x)^2/2(y/x)$

I am pretty sure that this is correct so far. Then,
$x(dv/dx)+v=1+3(y/x)^2/2(y/x)$
$x(dv/dx)=1+3(y/x)^2/2(y/x)-v$
$x(dv/dx)=(1+3v^2/2v)-v$

It is after this where I'm not quite sure what to do. According to my book the answer is $x^2+y^2-cx^3=0$. Thanks for any help
• Feb 1st 2010, 01:40 PM
Jester
Quote:

Originally Posted by steph3824
Show that the given equation is homogeneous and solve the differential equation.

$dy/dx=x^2+3y^2/2xy
$

To show that it is homogeneous, I have $dy/dx=1+3(y/x)^2/2(y/x)$

I am pretty sure that this is correct so far. Then,
$x(dv/dx)+v=1+3(y/x)^2/2(y/x)$
$x(dv/dx)=1+3(y/x)^2/2(y/x)-v$
$x(dv/dx)=(1+3v^2/2v)-v$

It is after this where I'm not quite sure what to do. According to my book the answer is $x^2+y^2-cx^3=0$. Thanks for any help

It helps to use brackets. So

$
\frac{dy}{dx} = \frac{x^2+3y^2}{2xy}
$

so if $y = x v$ then $\frac{dy}{dx} = x \frac{dv}{dx} + v$

so $x \frac{dv}{dx} = \frac{1 + 3v^2}{2v} - v = \frac{1+v^2}{2v}$ as you've said. Now separate

$
\frac{2v\,dv}{1+v^2} = \frac{dx}{x}
$
and integrate.
• Feb 1st 2010, 08:50 PM
steph3824
Ok I integrated that and ended up with $ln|1+v^2|=ln|x|+C$

Is that correct? Where do I go from here?
• Feb 2nd 2010, 05:04 AM
Calculus26
exponentiate

1+ v^2 = C x

replace v = y/x

simplify
• Feb 2nd 2010, 08:24 AM
Jester
Quote:

Originally Posted by steph3824
Ok I integrated that and ended up with $ln|1+v^2|=ln|x|+C$

Is that correct? Where do I go from here?

Quote:

Originally Posted by Calculus26
exponentiate

1+ v^2 = C x

replace v = y/x

simplify

Just an added note. As $C$ is constant in $ln|1+v^2|=ln|x|+C$, replace it with $\ln C$ and then follow Calc26.