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Math Help - First Order Differential Equation

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    First Order Differential Equation

    Question:
    \frac{dy}{dt} = \frac{t+y}{t+y+1} Solve for y(t)=...

    Approach:
    Letting u=t+y+1 then getting \frac{dy}{dt} = \frac{du}{dt}-1 and substituting into main equation to get \frac{du}{dt}-1 = \frac{u-1}{u} and rearraning to get \frac{du}{dt} + \frac{1}{u} = 2 however it's not in integration factor form. How would I solve it?
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  2. #2
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    Quote Originally Posted by Simplicity View Post
    Question:
    \frac{dy}{dt} = \frac{t+y}{t+y+1} Solve for y(t)=...

    Approach:
    Letting u=t+y+1 then getting \frac{dy}{dt} = \frac{du}{dt}-1 and substituting into main equation to get \frac{du}{dt}-1 = \frac{u-1}{u} and rearraning to get \frac{du}{dt} + \frac{1}{u} = 2 however it's not in integration factor form. How would I solve it?
    Try
    \frac{du}{dt} = 2 - \frac{1}{u} = \frac{2u-1}{u}

    It's separable.
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    Quote Originally Posted by danny arrigo View Post
    Try
    \frac{du}{dt} = 2 - \frac{1}{u} = \frac{2u-1}{u}

    It's separable.
    As by:

    \int \frac{u}{2u-1} du = \int dt?
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    Quote Originally Posted by Simplicity View Post
    As by:

    \int \frac{u}{2u-1} du = \int dt?
    Yes.
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    Quote Originally Posted by Mush View Post
    Yes.
    What method am I using to solve \int \frac{u}{2u-1} du?
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  6. #6
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    Quote Originally Posted by Simplicity View Post
    What method am I using to solve \int \frac{u}{2u-1} du?
     p = 2u-1

     dp/du = 2

    \frac{1}{2}dp = du

     \frac{1}{2} \int \frac{u}{p}dp

     u = \frac{p+1}{2}


     \frac{1}{2} \int \bigg(\frac{\frac{p+1}{2}}{p}\bigg)dp

     \frac{1}{4} \int\bigg(\frac{p+1}{p}\bigg)dp

     \frac{1}{4} \int \bigg(1+\frac{1}{p}\bigg) dp
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