Exact Equation with Integrating Factor Help

Homework Statement

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

Homework Equations

ydx+(2x-yey)dy=0, $$\mu$$(x,y)=y.

The Attempt at a Solution

M=y2, N=2xy-y2ey

Integrating N=$$\Psi$$y WRT x I get

xy2-((1/3)y3ey + y2ey)+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)y3ey + y2ey)

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

Homework Helper
Gold Member

Homework Statement

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

Homework Equations

ydx+(2x-yey)dy=0, $$\mu$$(x,y)=y.

The Attempt at a Solution

M=y2, N=2xy-y2ey

Integrating N=$$\Psi$$y WRT x I get

But you are looking for a function ψ such that ψx = M and ψy = N.

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

Staff Emeritus
Homework Helper
Gold Member
This is my take on this.

Multiplying by the integrating factor gives: y2dx+(2xy-y2ey)dy=0

y2dx/dy+2xy=y2ey

(d/dy)(xy2)=y2ey

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=Ψy(x,y) → 2xy+h'(y)=2xy-y2ey

h'(y) = -y2ey

Integrate this & plug it back into Ψ.

Last edited: