Initial Value Problem?

Question

I worked my way through this problem and ended up with 0 = 1/2.... which is not correct. =P

The problem is:
y y' = (e^2 - 1 + y^2) xe^x^2

And the condition is y(0) = 1.

I know you have to change y' to dy/dx, and then get all the y's on one side and all the x's on the other, so I got:

(ydy) / (e^2 - 1 + y^2) = xe^x^2 dx

I'm thinking I went wrong when I tried to integrate it, specifically on the left-hand side.

xe^x^2 integrated is (e^x^2) / 2, right?

And on the left, I got (y^2 / 2)(x / e^2 - 1 + y^2), which is where I have a feeling I went wrong.

Any help is appreciated, thanks!

Answer 1

To integrate (ydy) / (e^2 - 1 + y^2) , realize that if you let the denominator be u, you have (1/2)du in the numerator, so the anti-derivative is (1/2)ln|u|, or (1/2)ln(e^2 - 1 + y^2), removing the absolute value operation since this expression is always positive. For your right hand side I get the same answer.

--charlie

Answer 2

I plugged the integration into maple for you but didnt do the initial conditions.

for (ydy) / (e^2 - 1 + y^2) maple came up with
1/2 ln(e^2 - 1 + y^2)

for xe^x^2 maple came up with
1/2 ((e^x^2)/(ln(e))

hope this helps a little