How can I solve this intial value problem?

Question

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.

Answer 1

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

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

Answer 2

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