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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.

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2 Answers

up vote 8 down vote accepted

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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@Rasholnikov : Yeah this makes sense, but could you please explain how you divided every variable where it belongs before the second step? –  Binarylife Jun 14 '11 at 19:57
    
@binarylife: is this more like it? –  Raskolnikov Jun 14 '11 at 20:03
    
Yes this is very helpful , Thank you so much ! , I was stuck. –  Binarylife Jun 14 '11 at 20:07
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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. –  Binarylife Jun 14 '11 at 20:13
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