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)

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)

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New Wilmington, PA

The equation is separable. Separate and get

y dy = -x e^{-x²} dx

Integrate:

y^{2} = e^{-x²} + c

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

4 = 1+c ⇒ c=3 ⇒ y^{2} = e^{-x²} + 3

Since y>0,

y = sqrt(e^{-x²} + 3).

y = sqrt(e

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