Solve the differential equation dy/dx = -xe^y

Question

and determine the equation of the slope that passes by the point P(0,1)

I get to: ∫ 1/e^y dy = - ∫ x dx
which brings me to: -1/e^y + C = -x^2/2 + C?
What do you do after?

Thank you.

Answer 1

so far, so good
But you don't need a +C on each side of the equation. They can be combined into a different C, since C is just an arbitrary constant. So that leaves you with
-e^-y = -x^2/2 + C
e^-y = x^2/2 + C (a different C, just - the original C)
Now take logs
-y = log(x^2/2 + C)
But that's the same as -y = log (x^2+C)/2 ( a new C)
y = log 2/(x^2+C)

Answer 2

Thank you very much, I get it now.