Linear DE help.

I've got a linear DE here,

[tex] (x + 4y^2) dy + 2y dx =0 [/tex]

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

[tex] \frac {dy}{dx} + \frac {2y}{x+4y^2} = 0 [/tex]

but I have problems isolating the x, so that I would get the [tex] P(x) [/tex] in the general form.
 
Last edited:

AKG

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This doesn't make sense. If "DE" stands for "differential equation" then no, you don't have a DE here.
 
^ sorry, its supposed to be equals to 0, i forgot to add that...
 

Hurkyl

Staff Emeritus
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Think outside the box for a moment. There's a very easy way to turn this into an ODE.
 

James R

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Your D.E. is not linear, either.
 

HallsofIvy

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The way you have written it
[tex] \frac {dy}{dx} + \frac {2y}{x+4y^2} = 0 [/tex]
it is not linear.
However, you can write
[tex] (x + 4y^2) dy + 2y dx =0 [/tex]

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

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