Exact Equation with Integrating Factor Help

[Bogus_Roads]

Homework Statement

Multiply the given equation by the given integrating factor and solve the exact equation.

Homework Equations

ydx+(2x-ye^y)dy=0, \mu(x,y)=y.

The Attempt at a Solution

M=y², N=2xy-y²e^y

Integrating N=\Psi_y WRT x I get

xy²-((1/3)y³e^y + y²e^y)+h(x)=\Psi(x,y)

Differentiating \Psi(x,y) WRT x, I get

\Psi_x=y²+h'(x)

Thus h'(x)=0, and

\Psi(x,y)=xy²-((1/3)y³e^y + y²e^y)

The correct answer is xy²-(y²-2y+2)e^y=c...

What am I doing wrong? When I solve for psi in the opposite way I get the same wrong answer from before...

Thanks in advance!

 

[LCKurtz]

Homework Statement

Multiply the given equation by the given integrating factor and solve the exact equation.

Homework Equations

ydx+(2x-ye^y)dy=0, \mu(x,y)=y.

The Attempt at a Solution

M=y², N=2xy-y²e^y

Integrating N=\Psi_y WRT x I get

But you are looking for a function ψ such that ψ_x = M and ψ_y = N.

So to find ψ by integrating ψ_y = N, you need to integrate both sides with respect to y, not x.

 

[SammyS]

This is my take on this.

Multiplying by the integrating factor gives: y²dx+(2xy-y²e^y)dy=0

y²dx/dy+2xy=y²e^y

(d/dy)(xy²)=y²e^y

Integrate both sides WRT y.

\int \frac{d}{dy}(xy^2)\ dy=\int y^2\,e^y\, dy

So, to do it your way,

Integrate M=y² WRT x → Ψ(x,y)=xy²+h(y)

 → N=Ψ_y(x,y) → 2xy+h'(y)=2xy-y²e^y

∴ h'(y) = -y²e^y

Integrate this & plug it back into Ψ.

 

[Bogus_Roads]

Great, thanks for the help!