Homogeneous equation

**steph3824** · February 1st 2010, 12:05 PM

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

**Danny** · February 1st 2010, 12:40 PM

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.

**steph3824** · February 1st 2010, 07:50 PM

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

Is that correct? Where do I go from here?

**Calculus26** · February 2nd 2010, 04:04 AM

exponentiate

1+ v^2 = C x

replace v = y/x

simplify

**Danny** · February 2nd 2010, 07:24 AM

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?

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.

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