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Math Help - Differential question

  1. #1
    Junior Member superman69's Avatar
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    Differential question

    Solve the following initial value problem: with


    So I isolated the y and it became

    8y=(9-dy/dt)t

    how would I finish the rest?

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    Quote Originally Posted by superman69 View Post
    Solve the following initial value problem: with


    So I isolated the y and it became

    8y=(9-dy/dt)t

    how would I finish the rest?

    Divide both sides by t: \frac{dy}{dt} + \frac{8}{t} y = 9.

    Now use the integrating factor technique.
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  3. #3
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    You can also do this as a "homogeneous" equation. Write it as \frac{dy}{dt}= \frac{9t- 8y}{t}= 9- 8\frac{y}{t}.

    Let u= \frac{y}{t}. Then y= tu so y'= tu'+ u and the equation becomes tu'+ u= 9- 8u and so tu'= 9- 9u, a separable equation.
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    Junior Member superman69's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Divide both sides by t: \frac{dy}{dt} + \frac{8}{t} y = 9.

    Now use the integrating factor technique.
    What is the integrating factor technique? I did not learn that yet.
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  5. #5
    Junior Member superman69's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    You can also do this as a "homogeneous" equation. Write it as \frac{dy}{dt}= \frac{9t- 8y}{t}= 9- 8\frac{y}{t}.

    Let u= \frac{y}{t}. Then y= tu so y'= tu'+ u and the equation becomes tu'+ u= 9- 8u and so tu'= 9- 9u, a separable equation.
    So now I use the equation tu'= 9- 9u and plug in u' and u?
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    Quote Originally Posted by superman69 View Post
    What is the integrating factor technique? I did not learn that yet.
    \frac{dy}{dt} + \frac{8}{t}y = 9.


    The Integrating Factor is e^{\int{\frac{8}{t}\,dt}} = e^{8\ln{t}}

     = e^{\ln{t^8}}

     = t^8.


    So multiply both sides by the integrating factor to get

    t^8\left(\frac{dy}{dt} + \frac{8}{t}y\right) = 9t^8

    t^8\,\frac{dy}{dt} + 8t^7y = 9t^8


    Can you see that the LHS is a product rule expansion of \frac{d}{dt}(t^8y)?


    \frac{d}{dt}(t^8y) = 9t^8

    t^8y = \int{9t^8\,dt}

    t^8y = t^9 + C

    y = t + Ct^{-8}.


    Now use the initial condition to find C.
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  7. #7
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    Quote Originally Posted by superman69 View Post
    So now I use the equation tu'= 9- 9u and plug in u' and u?
    The equation is t \frac{du}{dt} = 9 - 9u. It's seperable and your task is to solve for u and so get y from y = ut (and I hope you have learnt about seperable DE's).
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    Quote Originally Posted by mr fantastic View Post
    The equation is t \frac{du}{dt} = 9 - 9u. It's seperable and your task is to solve for u and so get y from y = ut (and I hope you have learnt about seperable DE's).
    It helps if you write the DE as

    \frac{1}{1 - u}\,\frac{du}{dt} = \frac{9}{t}

    so that when you take the integral of both sides with respect to t

    \int{\frac{1}{1 - u}\,\frac{du}{dt}\,dt} = \int{\frac{9}{t}\,dt}

    \int{\frac{1}{1 - u}\,du} = \int{\frac{9}{t}\,dt}.


    I'm sure you can go from here.
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