# 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)

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

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

## Answers and Replies

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LCKurtz
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.

SammyS
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:
Great, thanks for the help!