What's f(x)?

Question

If the graph of y=f(x) contains the point (0, 2), dy/dx=-x/(ye^x^2) and f(x)>0 for all x, what's f(x)?

 

I integrated and got c=2. But I have trouble for finding y. Please show all the work.

 

Answer: sqrt(3+e^-x^2)

Answer 1

The equation is separable. Separate and get

y dy = -x e^-x² dx

Integrate:

y² = e^-x² + c

Impose the initial condition, y(0)=2:

4  = 1+c ⇒ c=3 ⇒ y² = e^-x² + 3

Since y>0,
y = sqrt(e^-x² + 3).