Solve (ye^xy+1)dx+(xe^xy+2)dy=0,y(0)=0

Answer 1

(ye^xy + 1)dx + (xe^xy + 2)dy = 0
Check for exactness:
Let M = ye^xy + 1
Let N = xe^xy + 2
∂M / ∂y = ∂N / ∂x
∂(ye^xy + 1) / ∂y = ∂(xe^xy + 2) / ∂x
Use chain rule to get the derivative:
e^xy + xy e^xy = e^xy + xy e^xy
Thus it is indeed exact.
To further solve,
∂F/∂x = M = ye^xy + 1
∫ ∂F = ∫ (ye^xy + 1)∂x
F = e^xy + x + g(y)

To get g(y), we differentiate it partially with respect to y:
∂F/∂y = ∂/∂y(e^xy + x + g(y)) = N
∂F/∂y = x e^xy + g'(y) = N
x e^xy + g'(y) = x e^xy + 2
g'(y) = 2
Integrating,
g(y) = 2y
Therefore,
F = e^xy + x + 2y

hope this helps~ `u`

Answer 2

Sorry I forgot to put the constant from integration. Thus,
F = e^xy + x + 2y + C