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