Question

e^(-y)dx-(2y+xe^(-y))dy=0

Answer 1

\(e^{-y}dx-(2y+xe^{-y})dy=0\)

doesn't look exact

Answer 2

thats the problem

Answer 3

Why not?
\(M(x,y) = e^{-y}\\M_y= -e^{-y}\)

\(N(x,y) = -2y-xe^{-y}\\N_x=-e^{-y}\)

Then, \(M_y=N_x\), it is exact.

Answer 4

cos its

\(d\phi = 0 = \phi_x dx + \phi_y dy \)

and with
\(e^{-y}dx-(2y+xe^{-y})dy=0\)

we have

\(\phi_x = e^{-y}, \phi_{xy} = -e^{-y}\)

\(\phi_y = -(2y+xe^{-y}), \phi_{yx} = -e^{-y}\)

so it is exact

Answer 5

yeah i got that. I'm stuck after that

Answer 6

\(\phi_x= e^{-y} \implies \phi = x e^{-y} + f(y)\)


\(\phi_y = -2y-xe^{-y} \implies \phi = -y^2 + x e^{-y} + g(x)= x e^{-y} -y^2 + g(x)\)


\(\phi = x e^{-y} -y^2 + C(x,y)\)

Answer 7

ask @Loser66 if that is right