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What's f(x)?

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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1 Answer

The equation is separable. Separate and get
y dy = -x e-x² dx
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).