# Solving dy/dx = e^(x+y)

• Jan 5th 2012, 03:14 PM
M.R
Solving dy/dx = e^(x+y)
Hi,

I am trying to solve the following differential equation:

$\displaystyle \frac {dy}{dx} = e^{x+y}$

Now:

$\displaystyle \frac {dy}{dx} = e^x e^y$

$\displaystyle \frac {1}{e^y} dy = e^x dx$

$\displaystyle \int \frac {1}{e^y} dy = \int e^x dx$

$\displaystyle -e^{-y} = e^x + C$

$\displaystyle ln(e^{-y^{-1})} = ln(e^x + C)$

$\displaystyle \frac{-1}{y} = x + ln(C)$

$\displaystyle y = \frac {1}{-x - ln(C)}$

But the answer in the book is shown as:

$\displaystyle e^{x+y}+Ce^y+1=0$

Where am I going wrong?
• Jan 5th 2012, 03:22 PM
pickslides
Re: Differential equation
Maybe you are correct, at this step $\displaystyle -e^{-y} = e^x + C$ , multiply both sides through by $\displaystyle e^{y}$

What do you get?
• Jan 5th 2012, 03:24 PM
ILikeSerena
Re: Differential equation
Quote:

Originally Posted by M.R
$\displaystyle ln(e^{-y^{-1})} = ln(e^x + C)$

$\displaystyle \frac{-1}{y} = x + ln(C)$

This is not right.
ln(ab) = ln(a) + ln(b)
ln(a + b) ≠ ln(a) + ln(b)

Quote:

Originally Posted by M.R
$\displaystyle -e^{-y} = e^x + C$

What do you get if you multiply left and right with $\displaystyle e^y$?