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Thread: Differential Equation

  1. #1
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    Differential Equation

    Can someone help me with this question?

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

    $\displaystyle \frac{d^2x}{dt^2} + 2 \frac{dx}{dt} + x = e^{-t}$ given that x = 1 and $\displaystyle \frac{dx}{dt} = 0$ 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 $\displaystyle x = At^{2}e^{-t}$ as a trial solution solve the differential equation ...

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


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

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

    You're given that the particular solution has the form $\displaystyle x = At^{2}e^{-t}$ (you should think about why that is) so you need to get the value of A. Start by calculating $\displaystyle \frac{dx}{dt}$ and $\displaystyle \frac{d^2x}{dt^2}$. 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 $\displaystyle \frac{dx}{dt} = 0$ 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 $\displaystyle x = At^{2}e^{-t}$

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


    therefore $\displaystyle At^{2}e^{-t} = e^{-t}$

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

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

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

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


    x = Particular Integral + Complimentary Function

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

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

    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 $\displaystyle x = At^{2}e^{-t}$

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

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


    therefore $\displaystyle At^{2}e^{-t} = e^{-t}$

    comparing coefficients:
    $\displaystyle At^{2} = 1$
    $\displaystyle 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 $\displaystyle x = ({\color{red}a} + Bt) e^{-{\color{red}t}}$. 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]
    $\displaystyle x = At^2 e^{-t}$

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

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

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

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

    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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