# Differential Questions

• Oct 21st 2009, 06:45 PM
electricalphysics
Differential Questions
Solve:

$\displaystyle \frac{dy}{dx}= \frac{y^3}{1-2xy^2},\ \ y(0)=1$

No idea how to do this, tried various methods and came up with nothing. Any Suggestions on how to attempt this one?
• Oct 22nd 2009, 06:14 AM
Jester
Quote:

Originally Posted by electricalphysics
Solve:

$\displaystyle \frac{dy}{dx}= \frac{y^3}{1-2xy^2},\ \ y(0)=1$

No idea how to do this, tried various methods and came up with nothing. Any Suggestions on how to attempt this one?

If you re-write your ODE as

$\displaystyle \frac{dx}{dy} = \frac{1-2xy^2}{y^3}$

or

$\displaystyle \frac{dx}{dy} + \frac{2x}{y} = \frac{1}{y^3}$

it's linear in $\displaystyle x$.
• Oct 22nd 2009, 07:41 AM
Krizalid
or divide top and bottom by $\displaystyle y^3$ and put $\displaystyle y=tx$ to turn your equation into a separable one.
• Oct 22nd 2009, 07:58 AM
electricalphysics
Quote:

Originally Posted by Danny
If you re-write your ODE as

$\displaystyle \frac{dx}{dy} = \frac{1-2xy^2}{y^3}$

or

$\displaystyle \frac{dx}{dy} + \frac{2x}{y} = \frac{1}{y^3}$

it's linear in $\displaystyle x$.

Thanks i got it, but
When am I to know I must flip the dy/dx to make x(y)? Is it because of the particular form of the ODE ?
• Oct 22nd 2009, 08:25 AM
HallsofIvy
Quote:

Originally Posted by electricalphysics
Thanks i got it, but
When am I to know I must flip the dy/dx to make x(y)? Is it because of the particular form of the ODE ?

When it works! Many mathematics problems are solved by "try this and if it doesn't work try that". The important thing is to try.