Which Differential Equation Method Do I use?

**lanczlot** · May 22nd 2010, 11:14 PM

i need to solve this differential equation (3y^3-xy)dx-(x^2+6xy^2)dy=0

I tried using Bernoulli, but it doesn't work. Also tried separating the variables but i cant seem to separate it. i also tried using the 3 cases for exact equation but to no avail. the 3rd case may work but i don't know how to simplify it so i can integrate it.

**mr fantastic** · May 23rd 2010, 12:09 AM

Originally Posted by lanczlot

i need to solve this differential equation (3y^3-xy)dx-(x^2+6xy^2)dy=0

I tried using Bernoulli, but it doesn't work. Also tried separating the variables but i cant seem to separate it. i also tried using the 3 cases for exact equation but to no avail. the 3rd case may work but i don't know how to simplify it so i can integrate it.

Are you sure it's not (3y^2 - xy) dx -(x^2 + 6xy^2) dy=0?

**lanczlot** · May 23rd 2010, 05:29 AM

the problem was hand written, so it may have been a mistake. what if it was squared instead of cube? how do you solve it?

**mr fantastic** · May 23rd 2010, 05:34 AM

Originally Posted by lanczlot

the problem was hand written, so it may have been a mistake. what if it was squared instead of cube? how do you solve it?

Divide through by y^2. Now read this: Homogeneous Ordinary Differential Equation -- from Wolfram MathWorld

The technique will be in your classnotes and textbook.

**Danny** · May 23rd 2010, 08:07 AM

Originally Posted by lanczlot

i need to solve this differential equation (3y^3-xy)dx-(x^2+6xy^2)dy=0

I tried using Bernoulli, but it doesn't work. Also tried separating the variables but i cant seem to separate it. i also tried using the 3 cases for exact equation but to no avail. the 3rd case may work but i don't know how to simplify it so i can integrate it.

If you write the ODE as

$\frac{dy}{dx} = \frac{3y^3-xy}{x^2 + 6xy^2}$

multiply both side by 2y

$2y\frac{dy}{dx} = \frac{6y^4-2xy^2}{x^2 + 6xy^2}$

and let $u = y^2$ then

$\frac{du}{dx} = \frac{6u^2-2xu}{x^2 + 6xu}$ (this is homogeneous).

**lanczlot** · May 23rd 2010, 09:53 AM

wait... how is the differential equation with u homogeneous?

**Danny** · May 23rd 2010, 11:55 AM

Divide everything on the rhs by $x^2$ so

$<br /> \frac{du}{dx} = \frac{6\dfrac{u^2}{x^2}-2\dfrac{u}{x}}{1 + 6\dfrac{u}{x}}$

which of the form is $\frac{du}{dx} = f\left(\frac{u}{x}\right)$ .

**Krizalid** · May 23rd 2010, 12:23 PM

the ODE also admits an integrating factor of the form $u(x,y)=x^my^n.$

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