Homogeneous Equation Help

• Feb 8th 2014, 11:50 PM
petenice
Homogeneous Equation Help
LaTex appears to not be working so ill type normally.

Should and can the equation below be solved using the y'=v+xv', y=vx substitution method? I try to do so and i get stuck halfway through.

-2y'+(4y/x)=y^3/x

-2(v+xv')+(4vx/x)=(v^3x^3)/x

-2v-2xv'+4v=v^3x^2

-2xv'+2v=v^3x^2

-2xv'=v^3x^2-2v

v'=-(v^3x^2-2v)/2x

Thats where i get stuck. Ive tried other ways but i still get stuck.
Thanks in advance for the help.
• Feb 9th 2014, 03:29 AM
Prove It
Re: Homogeneous Equation Help
It's a Bernoulli Equation.
• Feb 9th 2014, 11:17 PM
petenice
Re: Homogeneous Equation Help
Ok thank you. I managed to figure that one out. Took me a while but thanks for your input. But now im stuck at another question.

$\displaystyle y'=\frac{x^4y^4+4x^4}{y^3}$

So I tried to check if it was homogenous using $\displaystyle f(tx,ty)=f(x,y)$ and I concluded its not homogenous. Is that right?

I attempted to rearrange it and work it out as a Bernoulli equation but still I didnt get to the correct answer. Any guidance to which method I should be trying?
• Feb 10th 2014, 01:34 AM
Prove It
Re: Homogeneous Equation Help
If you divide each term by y^3 you'll find this one is also a Bernoulli equation.
• Feb 10th 2014, 02:21 PM
Soroban
Re: Homogeneous Equation Help
Hello, petenice!

Quote:

$\displaystyle y'\:=\:\frac{x^4y^4+4x^4}{y^3}$

The variables are Separable!

$\displaystyle \frac{dy}{dx} \;=\; \frac{x^4(y^4+4)}{y^3} \quad\Rightarrow\quad \frac{y^3}{y^4+4}\,dy \;=\;x^4\,dx$
• Feb 11th 2014, 07:10 PM
petenice
Re: Homogeneous Equation Help
I was able to figure it out once you guys set me in the right direction. I was making simple mistakes in the algebra. Thanks for the help everyone.