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Thread: Differential Equations

  1. #1
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    Question Differential Equations

    I am not sure how to do this problem. Can you help?

    Solve (t^2 + 1)dy/dx = yt-y.
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  2. #2
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    Is t a constant or did you mean $\displaystyle \frac{dy}{dt}$?
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  3. #3
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    Quote Originally Posted by graceofayak View Post
    I am not sure how to do this problem. Can you help?

    Solve (t^2 + 1)dy/dx = yt-y.
    $\displaystyle (t^2+1)y' = y(t-1)$

    $\displaystyle \frac{y'}{y} = \frac{t-1}{t^2+1}$ or $\displaystyle y=0$

    This is seperable. Solve.
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  4. #4
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    Diff. equation

    dy
    dt
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  5. #5
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    We are asked to find the value of $\displaystyle y$ in $\displaystyle (t^2+1)\frac{dy}{dt}=y(t-1)$.

    Like TPH said, this is a separable differential equation so

    $\displaystyle
    \frac{dy}{y} = \frac{t-1}{t^2+1}dt \Longleftrightarrow \int \frac{dy}{y} = \int \frac{t-1}{t^2+1}dt \Longleftrightarrow \ln{\left |y \right |} = \int \frac{t}{t^2+1}dt - \int \frac{dt}{t^2+1}
    $

    which gives us
    $\displaystyle
    \ln{\left |y \right |} = \frac{1}{2}\ln (t^2+1) - \arctan t + C
    $

    and solving for $\displaystyle y$, we arrive at

    $\displaystyle
    y=\frac{C\sqrt{t^2+1}}{e^{\arctan t}}
    $ for any constant C.

    We may verify that this is the general solution by differentiating $\displaystyle y$ and substituting it in the original differential equation.
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  6. #6
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    Thank you all for your help!!!!
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