# Constant

• Apr 6th 2010, 08:05 PM
cdlegendary
Constant
http://homework.math.ucsb.edu/webwor...76f81116a1.png

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
• Apr 6th 2010, 08:19 PM
pickslides
Quote:

Originally Posted by cdlegendary

Am I even going in the right direction? Thanks

(Nod)

Now integrate LHS by parts

$\displaystyle \int(y+1)e^{-y} ~dy = e^t+C$
• Apr 6th 2010, 08:28 PM
Prove It
Quote:

Originally Posted by cdlegendary
http://homework.math.ucsb.edu/webwor...76f81116a1.png

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 $\displaystyle C = C_1 - C_2$

$\displaystyle -t = \ln{\left[e^{-y}(y + 2) + C\right]}$

$\displaystyle t = -\ln{\left[e^{-y}(y + 2) + C\right]}$.