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How can I solve this IVP, 1st order differntial equation.

$\frac{dy}{dt}= \frac {1}{e^y-t}$

with initial value $y(1)=0?$

any help will be apperciated.

2 Answers 2

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By solving for the inverse function $t(y)$. Then the problem becomes

$\frac{dt}{dy} = e^y-t$

or

$\frac{dt}{dy} + t = e^y \; .$

Multiplying both sides by $e^y$

$e^y \frac{dt}{dy} + e^y t = e^{2y} \; ,$

and noting that the left hand side is the derivative of $e^y t$, we get

$\frac{d}{dy}\left( e^y t \right)= e^{2y} \; ,$

and integrating with respect to $y$ this becomes

$e^y t = \frac{1}{2}e^{2y} + C \; .$

Rearranging this, we arrive at

$t=\frac{1}{2}e^y-C e^{-y} \; .$

For your initial condition, this gives $t=\cosh(y)$ or

$y=\cosh^{-1}(t) \; .$

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    Yes this is very helpful , Thank you so much ! , I was stuck.2011-06-14
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Substituting $y = u + \ln t$ gives $ \frac{du}{dt} + \frac{1}{t} \;=\; \frac{1}{t e^u - t} $ which is a separable equation.

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    Thank you for the suggestion , I will try it though.2011-06-14