Consider the exact equation (exsiny+3y)dx+(excosy+3x)dy=0(e^x \sin y + 3y) dx + (e^x \cos y + 3x) dy = 0(exsiny+3y)dx+(excosy+3x)dy=0. Which function ψ(x,y)\psi(x,y)ψ(x,y) satisfies dψ=0d\psi = 0dψ=0?
ψ=exsiny+3xy=C\psi = e^x \sin y + 3xy = Cψ=exsiny+3xy=C
ψ=excosy+3xy=C\psi = e^x \cos y + 3xy = Cψ=excosy+3xy=C
ψ=exsiny−3xy=C\psi = e^x \sin y - 3xy = Cψ=exsiny−3xy=C
ψ=exsiny+3x+y=C\psi = e^x \sin y + 3x + y = Cψ=exsiny+3x+y=C