# Thread: Hard 1st order ODE Question

1. ## Hard 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!!!

2. Originally Posted by chemenger
(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!!!
Well:

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

so with some assumptions about smoothness of dy/dx you should be able to
proceed.

RonL

3. 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.

4. Originally Posted by chemenger
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.
Yes it should, unfortunatly with the y there its not so simple

RonL

5. Maybe your just have to approximate the solution.

Because it have the special form,
y'=f(x,y)
Apply Newton's Method.

6. Originally Posted by ThePerfectHacker
Maybe your just have to approximate the solution.

Because it have the special form,
y'=f(x,y)
Apply Newton's Method.
I can't. The question says solve this ode, which means I'm gonna need an analytical solution.

7. Originally Posted by chemenger
I can't. The question says solve this ode, which means I'm gonna need an analytical solution.
How about a series solution? (looks a bit tricky though)

RonL

8. I remember the lecturer said something about Laplace Transform, but I'm sure how I can relate that to this question.