Differential Equation Entry?

[DrD]

How can this be solved using the prime: (Ideally, returning BOTH a general and a particular solution.)

(y*e^x) dx + (y^2 - 1) dy = 0, AND y(0) = 1

Desired results:
y^2 - ln(y^2) = -2*e^x + c1; {General Solution}
y^2 - ln(y^2) = -2*e^x + 3; {Particular solution}