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Math Help - Initial Value Problem

  1. #1
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    Initial Value Problem

    Hello, thanks for clicking my thread.

    I need to solve the following IVP

    \dot{u} = (t+u)^2 ;

    u(0) = 0

    Does anyone know how to solve this?

    Kind regards
    Rapha
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  2. #2
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    Quote Originally Posted by Rapha View Post
    Hello, thanks for clicking my thread.

    I need to solve the following IVP

    \dot{u} = (t+u)^2 ;

    u(0) = 0

    Does anyone know how to solve this?

    Kind regards
    Rapha
    Ignore!
    Last edited by Mush; January 9th 2009 at 08:48 AM. Reason: Bollocks.
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  3. #3
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    Quote Originally Posted by Rapha View Post
    Hello, thanks for clicking my thread.

    I need to solve the following IVP

    \dot{u} = (t+u)^2 ;

    u(0) = 0

    Does anyone know how to solve this?

    Kind regards
    Rapha
    Let  v = t + u

    Hence  u = v - t

    And  LHS = \frac{du}{dt} = \frac{d(v-t)}{dt} = \frac{dv}{dt} - \frac{dt}{dt} =  \frac{dv}{dt} - 1

    And  RHS = v^2

    Hence \frac{dv}{dt} - 1= v^2

    \frac{dv}{dt} = v^2+1

    \frac{dv}{v^2+1} = dt

    \int \frac{dv}{v^2+1} = \int dt

     \arctan{(v)} = t+C^*

     v = \tan{(t+C^*)}

    Then using  \tan{(A+B)} = \frac{\tan{(A)}+\tan{(B)}}{1-\tan{(A)} \tan{(B)}} , we find that:

     v = \frac{\tan{(t)}+\tan{(C^*)}}{1-\tan{(t)} \tan{(C^*)}} = \frac{\tan{(t)}+C}{1-C\tan{(A)}} where  C = \tan{(C^*)}

    And remember that  u = v-t

     v-t= \frac{\tan{(t)}+C}{1-C\tan{(t)}} - t

     u= \frac{\tan{(t)}+C - (t-Ct\tan{(t)})}{1-C\tan{(t)}}

     u= \frac{\tan{(t)}+C - t+Ct\tan{(t)}}{1-C\tan{(t)}}

     u= \frac{\tan{(t)}(1+Ct)+C - t}{1-C\tan{(t)}}

    Now apply your initial value conditions to find C!
    Last edited by Mush; January 9th 2009 at 10:34 AM.
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  4. #4
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    See corrections in above post!
    Last edited by Mush; January 9th 2009 at 10:20 AM.
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  5. #5
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    Hi Mush.

    Thank you so much, you are great!


    Best wishes,
    Rapha
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