Differential equation

**ubique** · February 26th 2009, 08:16 PM

Hey guys i'm banging my head against the wall with this question.

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

i've tried multiplying top & bottom by e^-y and seperating, but I cant seem to match the answer in the back of the book which is e^x + ye^-y = C

The fact that the equation is equal to C gave me the idea that it might be exact, but when I tried the partial derivative test I couldn't get a single variable by itself (I might have screwed up there somewhere also)

Thanks in advance

**Soroban** · February 26th 2009, 08:36 PM

Hello, ubique!

You started off great . . .

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

i've tried multiplying top & bottom by $e^{-y}$ and separating . Good!
but I can't seem to match the answer in the book: . $e^x + ye^{-y} \:=\: C$

We have: . $\frac{dy}{dx} \:=\:\frac{e^x\cdot e^y}{y-1}$

Separate variables: . $\frac{y-1}{e^y}\,dy \;=\;e^x\,dx$ .[1]

On the left, we can integrate by parts:

. . $\begin{array}{ccccccc}u &=& y-1 & & dv &=& e^{-y}\,dy \\ du &=& dy & & v &=& \text{-}e^{-y} \end{array}$

And we have: . $-e^{-y}(y-1) + \int e^{-y}\,dy \;=\;-e^{-y}(y-1) - e^{-y}+C_1\;=\;-ye^{-y} +C_1$

Then [1] becomes: . $-ye^{-y} +C_1\;=\;e^x+ C_2 \quad\Rightarrow\quad\boxed{ e^x + ye^{-y} \:=\:C}$

**ubique** · February 26th 2009, 08:50 PM

Thanks a lot Soroban

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