Solve e^-x dy/dx = 1

when y=5 and x = 0

the solution I have been given is

this is far I go before getting confused as the solution I have been given is wrong or else my methods are wrong. can some correct me please?

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- Aug 3rd 2011, 06:03 AM #1

- Aug 3rd 2011, 06:13 AM #2
## Re: e^-x dy/dx = 1

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

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

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

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

$\displaystyle y=e^x+C$

$\displaystyle (0,5)$

$\displaystyle 5=e^0+C$

$\displaystyle 5=1+C$

$\displaystyle C=4$

Hence:

$\displaystyle y=e^x+4$

- Aug 3rd 2011, 06:31 AM #3
## Re: e^-x dy/dx = 1

You're integrating a term with x in as part of dy when it should be integrated with respect to x.

Since this equation is separable it's much easier to multiply both sides by $\displaystyle e^x$

$\displaystyle e^{-x} \cdot e^x\ dy = e^{x} \ dx \ \Leftrightarrow \ dy = e^x\ dx$

Then proceed as per Also sprach Zarathustra's method