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

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

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- Oct 19th 2011, 02:24 PMVitaXDifferential 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 PMSammySRe: Differential Equations - Integrating Trouble
- Oct 19th 2011, 02:45 PMVitaXRe: 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 PMVitaXRe: 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?