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Quick question I need help with.

Find the solution $x(t)$ satisfying initial value problem $\frac{dx}{dt} = e^x e^t$ where $x(0)=-3$.

After separating variables to get $-e^{-x}=e^t +c$, how do I get the equation in terms of $x$?

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up vote 1 down vote accepted

Your initial condition yields $$-e^3=1+c,$$ so $$c=-1-e^3,$$ whence $$-e^{-x}=e^t-1-e^3.$$ Then $$e^{-x}=-e^t+1+e^3,$$ so $$-x=\ln\left(-e^t+1+e^3\right),$$ and so $$x=-\ln\left(-e^t+1+e^3\right).$$

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Plugging your initial condition gives you $-e^{-x}=e^t-3$ or $e^{-x}=3-e^t$. You can then take the log to get $x=-\ln(3-e^t)$

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so when you plug in x(0)=-3 you just sub C for -3? and what happened to the $-e^-x$ why is it now $-e^x$ ? – Ricky Rozay Feb 26 '13 at 14:42
Yes, to use an initial condition you evaluate the equation at the time given, here $t=0$, and find the $c$ that makes the equation true. That is not always the value of $x$ at that time, but it is here. The loss of the sign in the exponent was an error that has been fixed. – Ross Millikan Feb 26 '13 at 14:45

You have the solution:-

$-\exp(-x) = \exp(t) + c$

To get the solution, put t=0 and x=-3 in the equation and it becomes:-

$-\exp(3) = \exp(0) +c$

or $-\exp(3)=1+c$

or $c=-1-\exp(3)$

So, the equation becomes :-

$-\exp(-x) = \exp(t) -1-\exp(3)$---------------(A)

or $ -\exp(-x) +1+\exp(3)=\exp(t)$

or $ t= \ln{\{-\exp(-x)+1+\exp(3)\}}$ where $\exp(x)=e^x$ (This gives you t in terms of x)

Continuing from (A), we can also say that:-

$\exp(-x) = -\exp(t) +1+\exp(3)$

or $-x=\ln\{-\exp(t)+1+\exp(3)\}$

or $x=-\ln\{-\exp(t)+1+\exp(3)\}$

or $x=\ln\frac{1}{\{1+\exp(3)-\exp(t)\}}$ where the symbols have their usual meaning.

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sorry, problem in writing LATEX. It is now rectified. – kusur Feb 27 '13 at 10:51

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