initial value problem:
$$y y'=x\textrm{e}^{-y^2}\qquad y(0)=2$$

Question

anonymous · Answer

$$y y'=x\textrm{e}^{-y^2}\qquad y(0)=2$$

anonymous · Answer

What type of equation is this?

anonymous · Answer

hell if I know, short of a differential equation.

anonymous · Answer

You want to do whats called separation of variables, im writing it out on paper, i'll post in a sec.

anonymous · Answer

so far ive got $$e^{y^2}=x^2$$

anonymous · Answer

$$y=\sqrt(2\ln(x))$$

anonymous · Answer

attachment

anonymous · Answer

Dang it, I always complicate things! I was thinking it was some kind of Bernoulli lol

anonymous · Answer

Forgot constants.  That is quite impressive you scan a handwritten working.

anonymous · Answer

Sometimes its faster for me than typing in LaTeX lol.

anonymous · Answer

I got $$C = e^4$$ and $$y = sqrt(2ln(x^2+e^4))$$

anonymous · Answer

incorrect

anonymous · Answer

i dont think that 2 belongs there.

vijay · Answer

$$\sqrt{\log(x^{2}} + e ^{4})$$

anonymous · Answer

lol you are correct.  I pulled it down from the log but then . . . didnt.  w/e answer is correct now. Thanks!