Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 8 down vote accepted

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

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


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

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

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

share|cite|improve this answer
@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

Substituting $y = u + \ln t$ gives $$ \frac{du}{dt} + \frac{1}{t} \;=\; \frac{1}{t e^u - t} $$ which is a separable equation.

share|cite|improve this answer
Thank you for the suggestion , I will try it though. – Binarylife Jun 14 '11 at 20:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.