Search
Ask a question
0

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)

1 Answer by Expert Tutors

Tutors, sign in to answer this question.
Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
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).