1. ## Homogeneous equation

Show that the given equation is homogeneous and solve the differential equation.

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

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

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

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

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

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

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

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

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

$\displaystyle \frac{dy}{dx} = \frac{x^2+3y^2}{2xy}$
so if $\displaystyle y = x v$ then $\displaystyle \frac{dy}{dx} = x \frac{dv}{dx} + v$

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

$\displaystyle \frac{2v\,dv}{1+v^2} = \frac{dx}{x}$ and integrate.

3. Ok I integrated that and ended up with $\displaystyle ln|1+v^2|=ln|x|+C$

Is that correct? Where do I go from here?

4. exponentiate

1+ v^2 = C x

replace v = y/x

simplify

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

Is that correct? Where do I go from here?
Originally Posted by Calculus26
exponentiate

1+ v^2 = C x

replace v = y/x

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