(dy/dx)^2 - 4x(dy/dx) + 6y = 0

The dy/dx squared term is kind of nasty. I've tried with all the substitutes I could think of, none of them worked.

Any hints would be appreciated. Thanks!!!

Printable View

- April 13th 2007, 07:03 AMchemengerHard 1st order ODE Question
(dy/dx)^2 - 4x(dy/dx) + 6y = 0

The dy/dx squared term is kind of nasty. I've tried with all the substitutes I could think of, none of them worked.

Any hints would be appreciated. Thanks!!! - April 13th 2007, 07:16 AMCaptainBlack
- April 13th 2007, 07:24 AMchemenger
First off, thanks for the quick reply:)

Shouldn't that be dy/dx = 2x +/- sqrt(4 x^2 -6y)

I think you were missing a y.

I've actually thought about using the quadratic eqn. But I wasn't sure how to deal with the stuff that's under the sqrt sign.

Could you make it a bit clear? Thanks again. - April 13th 2007, 07:32 AMCaptainBlack
- April 13th 2007, 08:36 AMThePerfectHacker
Maybe your just have to approximate the solution.

Because it have the special form,

y'=f(x,y)

Apply Newton's Method. - April 13th 2007, 04:57 PMchemenger
- April 13th 2007, 11:56 PMCaptainBlack
- April 14th 2007, 12:13 AMchemenger
I remember the lecturer said something about Laplace Transform, but I'm sure how I can relate that to this question.