First order differential.

dy/dx = xexp(y - x2)

dy/dx = xexp(y)/exp(x2)

By separating variables,

S e-y dy = S xe(-x^2) dx

-e-y = -1/2 S e(-x^2) d(-x2)

e-y = 1/2 e(-x^2) + C

When x = 0, y = 0

So, C = 1 - 1/2 = 1/2

Therefore, e-y = (e-x^2 + 1)/2

ey = 2/(e-x^2 + 1)

y = ln[]


2009-10-04 14:16:20 補充：
y = ln[2/(e-x^2 + 1)]