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

**M.R** · January 5th 2012, 03:14 PM

Hi,

I am trying to solve the following differential equation:

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

Now:

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

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

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

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

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

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

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

But the answer in the book is shown as:

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

Where am I going wrong?

**pickslides** · January 5th 2012, 03:22 PM

Maybe you are correct, at this step $-e^{-y} = e^x + C$ , multiply both sides through by $e^{y}$

What do you get?

**ILikeSerena** · January 5th 2012, 03:24 PM

Originally Posted by M.R

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

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

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

Originally Posted by M.R

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

You already had this.
What do you get if you multiply left and right with $e^y$ ?

**M.R** · January 5th 2012, 03:28 PM

Thank you

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