# Differential Equations - Integrating Trouble

• Oct 19th 2011, 02:24 PM
VitaX
Differential Equations - Integrating Trouble
$\displaystyle \int dy = \int \left[\frac{t}{1 - C_1 t}\right]$

I'm really at a loss on how to integrate that right hand side, what do I need to do here?
• Oct 19th 2011, 02:39 PM
SammyS
Re: Differential Equations - Integrating Trouble
Quote:

Originally Posted by VitaX
$\displaystyle \int dy = \int \left[\frac{t}{1 - C_1 t}\right]$

I'm really at a loss on how to integrate that right hand side, what do I need to do here?

Is the variable of integration t ?

$\displaystyle \int \left[\frac{t}{1 - C_1 t}dt\right]$

Let $\displaystyle u = 1 - (C_1) t\,,$ then $\displaystyle du = - (C_1)dt\,$ and $\displaystyle t=\frac{1-u}{C_1}$
• Oct 19th 2011, 02:45 PM
VitaX
Re: Differential Equations - Integrating Trouble
Yes the variable of integration is t and C1 is just some constant that I got integrating the problem earlier. It was originally a second order differential equation with the dependent variable missing. It's just I got down to this step and was confused on how to integrate it to finish it off. I'm a little confused on your work above, can you elaborate a little bit more? I thought a U substitution was useless here.
• Oct 19th 2011, 05:49 PM
VitaX
Re: Differential Equations - Integrating Trouble
$\displaystyle y(t) = \int \left[\frac{t}{1 - C_1 t}\right] dt$

$\displaystyle let: u = 1 - C_1 t$
$\displaystyle du = -C_1 dt \Longrightarrow \frac{-1}{C_1}du = dt$

$\displaystyle \frac{-1}{C_1} \int \left(\frac{1-u}{C_1}\right)\left(\frac{1}{u}\right) du$

$\displaystyle \frac{-1}{C_1} \int \left(\frac{1-u}{C_1 u}\right) du$

$\displaystyle \frac{-1}{C_1^2} \int \left(\frac{1-u}{u}\right) du$

$\displaystyle \int \left(\frac{-1 + u}{C_1^2 u}\right) du$

$\displaystyle \int \left(\frac{-1}{C_1^2 u}\right) du + \int \left(\frac{1}{C_1^2}\right) du$

Is this correct thus far?