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Math Help - Differential Equation

  1. #1
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    Differential Equation

    Can someone help me with this question?

    Using x = At^{2}e^{-t} as a trial solution solve the differential equation ...

     \frac{d^2x}{dt^2} + 2 \frac{dx}{dt} + x = e^{-t} given that x = 1 and \frac{dx}{dt} = 0 at t = 0.


    Use your solution to prove that for t>=0, x<=1

    Thanks in advance
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  2. #2
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    Quote Originally Posted by intomath View Post
    Can someone help me with this question?

    Using x = At^{2}e^{-t} as a trial solution solve the differential equation ...

     \frac{d^2x}{dt^2} + 2 \frac{dx}{dt} + x = e^{-t} given that x = 1 and \frac{dx}{dt} = 0 at t = 0.


    Use your solution to prove that for t>=0, x<=1

    Thanks in advance
    First solve  \frac{d^2x}{dt^2} + 2 \frac{dx}{dt} + x = 0 to get the homogenous solution. Can you do this?

    You're given that the particular solution has the form x = At^{2}e^{-t} (you should think about why that is) so you need to get the value of A. Start by calculating \frac{dx}{dt} and \frac{d^2x}{dt^2}. Can you do this?


    Note: Once you have the value of A and hence the particular solution you can construct the complete solution by adding together the homogenous and particular solutions. The boundary conditions x = 1 and \frac{dx}{dt} = 0 at t = 0 are used to get the value of the arbitrary constants.
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  3. #3
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    This is how i worked it out so far:

    Using trial solution x = At^{2}e^{-t}

    \frac{dx}{dt} = 0
    \frac{d^{2}x}{dt^{2}} = 0


    therefore At^{2}e^{-t} = e^{-t}

    comparing coefficients:
    At^{2} = 1
     t = \sqrt{\frac{1}{A}} ....Particular Integral

    Considering
     \frac{d^2x}{dt^2} + 2 \frac{dx}{dt} + x = 0

    Let  x = e^{ut} be a trial solution
    therefore  u^{2} + 2u + 1 = 0
    u = -1(twice) therefore real equal roots

    hence  x = (A+Bt)e^{-x} - Complimentary Function


    x = Particular Integral + Complimentary Function

    therefore  x = \sqrt{\frac{1}{A}} + (A+Bt)e^{-x} ....[A]

    putting x = 1, t = 0 in [A]
     1 =   \sqrt{\frac{1}{A}}  +    (A+B(0))e^{-1}

    This is where i'm stuck, i don't know how to carry on from here.
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  4. #4
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    Quote Originally Posted by intomath View Post
    This is how i worked it out so far:

    Using trial solution x = At^{2}e^{-t}

    \frac{dx}{dt} = 0
    \frac{d^{2}x}{dt^{2}} = 0

    Mr F says: I have no idea how you got the above results but they are wrong.


    therefore At^{2}e^{-t} = e^{-t}

    comparing coefficients:
    At^{2} = 1
     t = \sqrt{\frac{1}{A}} ....Particular Integral

    Mr F says: None of the above is correct.

    And there are various careless errors in your complementary solution as well: should be x = ({\color{red}a} + Bt) e^{-{\color{red}t}}. The pronumeral A should not be used in the complementary solution because this symbol is already used in the question to mean something else.



    [snip]
    x = At^2 e^{-t}

    \Rightarrow \frac{dx}{dt} = Ae^{-t} (2t - t^2)

    \Rightarrow \frac{d^2x}{dt^2} = Ae^{-t} (t^2 - 4t + 2).

    Carefully substitute the above into the DE and simplify. You will be left with 2A = 1 \Rightarrow A = \frac{1}{2}.

    So x = (a + Bt) e^{-t} + \frac{t^2}{2} e^{-t}.

    Now use the given boundary conditions x = 1 and at t = 0 to solve for a and B.
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  5. #5
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    Thanks a lot you saved my day
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