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Thread: Integrating Factor Problem

  1. #1
    Member VitaX's Avatar
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    Integrating Factor Problem

    Find the solution of the initial value problem.

    $\displaystyle ty' + (t+1)y = t$ ; $\displaystyle y(ln2) = 1$ ; $\displaystyle t > 0$

    $\displaystyle \frac{dy}{dx} + p(x)y = Q(x)$

    $\displaystyle y' + \left(1 + \frac{1}{t} \right)y = 1$

    $\displaystyle p(t) = 1 + \frac{1}{t}$ ; $\displaystyle Q(t) = 1$

    $\displaystyle \int p(x) = \int 1 + \frac{1}{t} dt = t + ln(t)$

    Integrating Factor: $\displaystyle e^{t + ln(t)} = e^{t}e^{ln(t)} = te^t$

    $\displaystyle te^t \left[y' + \left(1 + \frac{1}{t} \right)y \right] = te^t$

    $\displaystyle te^t y' + \left(1 + \frac{1}{t} \right)y = te^t$

    $\displaystyle u'v + v'u = \frac{d}{dx} (uv)$ ; $\displaystyle u = y$ ; $\displaystyle v = te^t$

    $\displaystyle \frac{d}{dt}[yte^t] = te^t$

    $\displaystyle \int d[yte^t] = \int te^t dt$

    $\displaystyle yte^t = \int te^t dt$

    Integrate by Parts: $\displaystyle \int te^t dt = te^t - e^t$

    $\displaystyle yte^t = te^t - e^t$

    $\displaystyle y(t) = 1 - \frac{1}{t} + C$

    Initial Condition: $\displaystyle y(ln2) = 1$

    $\displaystyle y(ln2) = 1 = 1 - \frac{1}{ln2} + C$

    $\displaystyle C = \frac{1}{ln2}$

    My Solution: $\displaystyle y(t) = 1 - \frac{1}{t} + \frac{1}{ln2}$

    Book Solution: $\displaystyle y(t) = \frac{t - 1 + 2e^{-t}}{t}$

    Where exactly did I go wrong?
    Last edited by VitaX; Sep 5th 2011 at 04:19 PM.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Re: Integrating Factor Problem

    Quote Originally Posted by VitaX View Post
    Find the solution of the initial value problem.

    $\displaystyle ty' + (t+1)y = t$ ; $\displaystyle y(ln2) = 1$ ; $\displaystyle t > 0$

    $\displaystyle \frac{dy}{dx} + p(x)y = Q(x)$

    $\displaystyle y' + \left(1 + \frac{1}{t} \right)y = 1$

    $\displaystyle p(t) = 1 + \frac{1}{t}$ ; $\displaystyle Q(t) = 1$

    $\displaystyle \int p(x) = \int 1 + \frac{1}{t} dt = t + ln(t)$

    Integrating Factor: $\displaystyle e^{t + ln(t)} = e^{t}e^{ln(t)} = te^t$

    $\displaystyle te^t \left[y' + \left(1 + \frac{1}{t} \right)y \right] = te^t$

    $\displaystyle te^t y' + \left(1 + \frac{1}{t} \right)y = te^t$

    $\displaystyle u'v + v'u = \frac{d}{dx} (uv)$ ; $\displaystyle u = y$ ; $\displaystyle v = te^t$

    $\displaystyle \frac{d}{dt}[yte^t] = te^t$

    $\displaystyle \int d[yte^t] = \int te^t dt$

    $\displaystyle yte^t = \int te^t dt$

    Integrate by Parts: $\displaystyle \int te^t dt = te^t - e^t$

    $\displaystyle yte^t = te^t - e^t$ This is where you made your mistake.

    $\displaystyle y(t) = 1 - \frac{1}{t} + C$

    Initial Condition: $\displaystyle y(ln2) = 1$

    $\displaystyle y(ln2) = 1 = 1 - \frac{1}{ln2} + C$

    $\displaystyle C = \frac{1}{ln2}$

    My Solution: $\displaystyle y(t) = 1 - \frac{1}{t} + \frac{1}{ln2}$

    Book Solution: $\displaystyle y(t) = \frac{t - 1 + 2e^{-t}}{t}$

    Where exactly did I go wrong?
    You made your mistake when you said $\displaystyle yte^t = te^t-e^t$. It should be $\displaystyle yte^t=te^t-e^t{\color{red} + C}$.

    That way, when you divide through by $\displaystyle te^t$, you get $\displaystyle y=1-\frac{1}{t}+\frac{1}{t}Ce^{-t}$

    Now apply your initial condition to get the book's solution.

    I hope this helps.
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  3. #3
    Member VitaX's Avatar
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    Re: Integrating Factor Problem

    Ah, OK. Thanks for clearing that up.
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