Uniqueness of Solution for Differential Equation with Initial Condition y(0) = 0

[tripleZ]

Homework Statement

Show that this problem has a unique solution:

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

given that y(0) = 0.

Homework Equations

Test for exactness: If (when rewritten into (2+2x^2)y' - 4x+2e^y = 0 ; which i hope is correct) My = Nx then there is an exact solution.

The Attempt at a Solution

I set M(x,y) = -4x+2e^y, and N(x,y) = 2+2x^2. (those can be further simplified by dividing by 2).
Then I found My(x,y) partial to be -2x +e^y, and Nx(x,y) = 2x.

From there on I couldn't think of much, I tried finding integrating factor but failed in that too. Would be nice if someone can point me in the right direction. I should also mention that I've had a break from maths for a few years (last I did was intro calculus) and now that I'm back I noticed that I've forgot a LOT, so I hope to get some help here to get me started :)

 

[PseudoIntellect]

It looks like e^-y works as an integrating factor. From there, it becomes closed and exact.

 

[tripleZ]

Thank you. Did you see that by just looking at it or calculate it using the integrating factor theorem?

I'm not having much luck in understanding how to get the integrating factor, my book has only 1 example and it doesn't help me much, how would you rewrite the original equation into the form y' + P(x)y = Q(x)?