Solving a Differential Equation: y`=1/(6e^y-2x)

[asdf1]

for the following question:
y`=1/(6e^y-2x)

my problem:
dx-(6e^y-2x)dy=0
so (1/F)(dF/dx)=2
=>F=e^2x
so e^2x-(e^2x)(6e^y-2x)dy=0
so (integration)(d^2x)dx=(1/2)e^2x +h(y)
=>h(y)= -6e^(2x+y)+xye^2x+c
so (1/2)e^2x-6e^(2x+y)+xye^2x+c-0

however the correct answer should be
y=ce^(-2y)+2e^y

does anybody know where my calculations went wrong?

 

[Fermat]

Your correct answer hasn't got an x-term in it. Is that a typo ?

 

[asdf1]

no, it's not a typo~

 

[Fermat]

Did a little research on this. Finally figured out what you were trying to do
You have a differential equation and you were trying to force it to be an exact DE, yes ?

I'm afraid you got the integrating factor wrong.
You had,

(1/F)dF/dx = 2,

that should have been

(1/F(y))dF(y)/dy = 2

giving,

F(y) = e^(3y)

Also, your "correct answer" is wrong,

$$y = ce^{-2y} + 2e^y$$

There is no x-term in this eqn, so you have no relation between x and y, so you can't use this eqn to get a DE that involves an x, since x was never involved in the first place.

The actual answer is,

$$c = x.e^{2y} - 2e^{3y}$$
or
$$x = c.e^{-2y} + 2e^y$$

If you differentiate that, you'll end with the DE you started with.

 

[dextercioby]

HINT:Make $e^{y(x)}=t(x)$

Daniel.

 

[asdf1]

opps... i corrected the integrating factor and it works! f=e^2y~
why did you think of making $e^{y(x)}=t(x)$?

 

[Fermat]

I guess that's a substitution you get to know with experience ?

It's faster/simpler than doing the exact differential thing.