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Math Help - Differential Equations - Integrating Trouble

  1. #1
    Member VitaX's Avatar
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    Differential Equations - Integrating Trouble

    \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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  2. #2
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    Re: Differential Equations - Integrating Trouble

    Quote Originally Posted by VitaX View Post
    \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 ?

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

    Let u = 1 - (C_1) t\,, then du =  - (C_1)dt\, and t=\frac{1-u}{C_1}
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  3. #3
    Member VitaX's Avatar
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    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.
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  4. #4
    Member VitaX's Avatar
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    Re: Differential Equations - Integrating Trouble

    y(t) = \int \left[\frac{t}{1 - C_1 t}\right] dt

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

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

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

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

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

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

    Is this correct thus far?
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