Results 1 to 8 of 8

Thread: Differential question

  1. #1
    Junior Member superman69's Avatar
    Joined
    Oct 2009
    From
    Hauts-de-Seine
    Posts
    30

    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?

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    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: $\displaystyle \frac{dy}{dt} + \frac{8}{t} y = 9$.

    Now use the integrating factor technique.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,730
    Thanks
    3011
    You can also do this as a "homogeneous" equation. Write it as $\displaystyle \frac{dy}{dt}= \frac{9t- 8y}{t}= 9- 8\frac{y}{t}$.

    Let $\displaystyle u= \frac{y}{t}$. Then $\displaystyle y= tu$ so y'= tu'+ u and the equation becomes tu'+ u= 9- 8u and so tu'= 9- 9u, a separable equation.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member superman69's Avatar
    Joined
    Oct 2009
    From
    Hauts-de-Seine
    Posts
    30
    Quote Originally Posted by mr fantastic View Post
    Divide both sides by t: $\displaystyle \frac{dy}{dt} + \frac{8}{t} y = 9$.

    Now use the integrating factor technique.
    What is the integrating factor technique? I did not learn that yet.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member superman69's Avatar
    Joined
    Oct 2009
    From
    Hauts-de-Seine
    Posts
    30
    Quote Originally Posted by HallsofIvy View Post
    You can also do this as a "homogeneous" equation. Write it as $\displaystyle \frac{dy}{dt}= \frac{9t- 8y}{t}= 9- 8\frac{y}{t}$.

    Let $\displaystyle u= \frac{y}{t}$. Then $\displaystyle y= tu$ 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?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by superman69 View Post
    What is the integrating factor technique? I did not learn that yet.
    $\displaystyle \frac{dy}{dt} + \frac{8}{t}y = 9$.


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

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

    $\displaystyle = t^8$.


    So multiply both sides by the integrating factor to get

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

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


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


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

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

    $\displaystyle t^8y = t^9 + C$

    $\displaystyle y = t + Ct^{-8}$.


    Now use the initial condition to find $\displaystyle C$.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    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 $\displaystyle t \frac{du}{dt} = 9 - 9u$. 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).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by mr fantastic View Post
    The equation is $\displaystyle t \frac{du}{dt} = 9 - 9u$. 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

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

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

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

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


    I'm sure you can go from here.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. differential question..
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 20th 2010, 08:05 AM
  2. A question about a differential
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 9th 2010, 07:07 AM
  3. differential question
    Posted in the Differential Equations Forum
    Replies: 7
    Last Post: Jul 5th 2009, 04:57 PM
  4. Differential question
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Jun 13th 2008, 02:46 PM
  5. Differential question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 25th 2007, 05:28 AM

Search Tags


/mathhelpforum @mathhelpforum