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misogynisticfeminist
Nov29-04, 09:38 PM
I've got a linear DE here,

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

I've tried to put it in the general form of a linear equation, and I would get,

\frac {dy}{dx} + \frac {2y}{x+4y^2} = 0

but I have problems isolating the x, so that I would get the P(x) in the general form.

AKG
Nov29-04, 09:45 PM
This doesn't make sense. If "DE" stands for "differential equation" then no, you don't have a DE here.

misogynisticfeminist
Nov29-04, 09:48 PM
^ sorry, its supposed to be equals to 0, i forgot to add that...

Hurkyl
Nov29-04, 10:05 PM
Think outside the box for a moment. There's a very easy way to turn this into an ODE.

James R
Nov30-04, 06:13 AM
Your D.E. is not linear, either.

HallsofIvy
Nov30-04, 07:03 AM
The way you have written it
\frac {dy}{dx} + \frac {2y}{x+4y^2} = 0
it is not linear.
However, you can write
(x + 4y^2) dy + 2y dx =0

as
\frac{dx}{dy}= -\frac{x+4y^2}{2y}= -\frac{1}{2y}x- 2y
which is a linear d.e. for x as a function of y.