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) Dec 24 | Sun from Los Angeles, CA | 1 Answer | 0 Votes Mark favorite Subscribe Comment
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). Dec 25 | Andre W. Comment