Find the particular solution of the differential equation ex√(1-y²)dx + yxdy = 0, given that y=1 when x=0.

Let I = ∫ex√(1-y²)dx = -∫yxdy

x exdx = -∫y√(1-y²)dy
Integrating both sides,
∫xexdx = 1/2∫-2y√(1-y²)dy
=> xex - ex = √(1-y²) + C
For x=0, y=1, we get
=> (0)e0 - e0 = √(1-(1)²) + C
=> C = -1
Therefore, xex - ex = √(1-y²) - 1