# Exact ODE

budafeet57

## Homework Statement

From Elementary Differential Equation by Boyce and Diprima
Chapter 2 Miscellaneous Problems #11
(x^2+y)dx + (x+e^x)dy = 0

ANS:(x^3/3)+xy+e^x=c

## Homework Equations

multiplying an integrating factor to make the DE exact:
1. du/dx = u(My - Nx)/ N

2. du/dx = u(Nx-My)/ M

## The Attempt at a Solution

First try: I guessed this can be changed into exact DE so, I tried with the two above equation:
equation 1 gave me:

du/u = e^x/(x+e^x)
I don't know how to solve this...

then equation 2 gave me:

u = e^((e^x)*ln(x^2+y))

I am not sure if multiply this integrating factor to the original DE will make it exact...

Second try: I manipulated the given DE and changed it to a linear form:

dy/dx = -(x^2+y)/(x+e^x)

dy/dx + 1/(x+e^x) * y = (-x^2)/(x+e^x)

and I found integrating factor to be:

I = e^∫1/(x+e^x) dx

which I am unable to solve...

budafeet57
Third try:

take integral of both side

∫(x^2+y)dx + (x+e^x)dy = ∫0

(x^3)/3 + yx + xy + ye^x = c //move constant c from left side to right side

(x^3)/3 + 2xy + ye^x = c

however it's not quite the same as the answer...

Wolframalfa does not solve this one!!!

voko
## (x^3/3)+xy+e^x=c ## cannot be the solution for ##(x^2+y)dx + (x+e^x)dy = 0##.

Differentiate: $$d((x^3/3)+xy+e^x) = d(x^3/3) + d(xy) + d(e^x) \\ = x^2dx + xdy + ydx + e^xdx \\ = (x^2 + y + e^x)dx + xdy \ne (x^2+y)dx + (x+e^x)dy$$