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.