I need to solve for all constant solutions for the equation above. How do I go about doing this?
$\displaystyle \int(y+1)/e^y dy = \int(e^t)dt$
Am I even going in the right direction? Thanks
$\displaystyle \frac{dy}{dt} = \frac{e^{t + y}}{y + 1}$
$\displaystyle (y + 1)\,\frac{dy}{dt} = e^te^y$
$\displaystyle e^{-y}(y + 1)\,\frac{dy}{dt} = e^t$
$\displaystyle \int{e^{-y}(y + 1)\,\frac{dy}{dt}\,dt} = \int{e^{-t}\,dt}$
$\displaystyle \int{e^{-y}(y + 1)\,dy} = -e^{-t} + C_1$.
Now use integration by parts:
Let $\displaystyle u = y + 1$ so that $\displaystyle du = 1$
Let $\displaystyle dv = e^{-y}$ so that $\displaystyle v = -e^{-y}$.
$\displaystyle -e^{-y}(y + 1) - \int{-1e^{-y}\,dy} = -e^{-t} + C_1$
$\displaystyle -e^{-y}(y + 1) + \int{e^{-y}\,dy} = -e^{-t} + C_1$
$\displaystyle -e^{-y}(y + 1) - e^{-y} + C_2 = -e^{-t} + C_1$
$\displaystyle -e^{-y}(y + 2) + C_2 = -e^{-t} + C_1$
$\displaystyle e^{-y}(y + 2) - C_2 = e^{-t} - C_1$
$\displaystyle e^{-y}(y + 2) + C = e^{-t}$ where $\displaystyle C = C_1 - C_2$
$\displaystyle -t = \ln{\left[e^{-y}(y + 2) + C\right]}$
$\displaystyle t = -\ln{\left[e^{-y}(y + 2) + C\right]}$.