Question

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

Answer 1

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

Answer 2

What type of equation is this?

Answer 3

hell if I know, short of a differential equation.

Answer 4

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

Answer 5

so far ive got \[e^{y^2}=x^2\]

Answer 6

\[y=\sqrt(2\ln(x))\]

Answer 7

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

Answer 8

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

Answer 9

Sometimes its faster for me than typing in LaTeX lol.

Answer 10

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

Answer 11

incorrect

Answer 12

i dont think that 2 belongs there.

Answer 13

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

Answer 14

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