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Math Help - find value in ODE

  1. #1
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    find value in ODE

    Hi , Please Can I have some help with this problem:

    Find the values of D such that the solution of the equation is bounded:

    y''-4y'= sin(x); y(0)=D, y'(0)=0.


    I started by applying the laplace Transform to the eqn.
    I found (if not mistakes)

    L(y)= [sD/(s^2-4) ] + [ 1/(s^2+1)(s^2-4)]

    After that I know that I need to focus on the first term with D but how to apply the condition for bounded??

    Thank you for your help

    B
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  2. #2
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    Quote Originally Posted by braddy
    Hi , Please Can I have some help with this problem:

    Find the values of D such that the solution of the equation is bounded:

    y''-4y'= sin(x); y(0)=D, y'(0)=0.


    I started by applying the laplace Transform to the eqn.
    I found (if not mistakes)

    L(y)= [sD/(s^2-4) ] + [ 1/(s^2+1)(s^2-4)]

    After that I know that I need to focus on the first term with D but how to apply the condition for bounded??

    Thank you for your help

    B
    I do not understand what "bounded" means. Anyways, let me solve the differencial equation maybe that would help.
    I think LaPlace Transforms is a primitive way to solve this.
    Just use the charachteristic equation.
    You have,
    y''-4y'=\sin x
    In such a case, there is a theorem. "The solution to a non-homogenenous equation is the sum of the general solution to the homogenous equation and a specific solution to the non-homogenous solution."

    First solve the homogenous equation:
    y''-4y'=0 thus, its charachteristic is,
    k^2-4k=0 thus, k=0,4. Thus, the general solutions to this homogenous equation are,
    y=C_1e^{0x}+C_2e^{4x}=C_1+C_2e^{4x}

    Second, find a specific solution to the non-homogenous,
    y''-4y'=\sin x. Since, there is a sine on the right look for solutions of the form,
    y=A\sin x+B\cos x thus,
    y'=A\cos x-B\sin x thus,
    y''=-A\sin x-B\cos x.
    Substitute that, into the non-homogenous
    <br />
-A\sin x-B\cos x-4A\cos x +4B\sin x=\sin x<br />
    Thus,
    (-A+4B)\sin x+(-B-4A) \cos x=\sin x
    Thus,
    \left\{\begin{array}{c}-A+4B=1\\-B-4A=0\end{array}\right
    Solving, we have (A,B)=(1/15,4/15)

    Now, by the theorem, all general solution to this non-homogenous equation are the sum of these two solutions. Which gives,
    y=C_1+C_2e^{4x}+\frac{1}{15}\sin x+\frac{4}{15}\cos x
    ----------
    Looking at the initial conditions: Evaluating this at zero we get D,
    C_1+C_2+\frac{4}{15}=D
    Its, derivative is when evaluated at zero is,
    4C_2+\frac{1}{15}=0
    Thus, solving these two equations,
    C_1=D-1/4
    C_2=-1/60

    Thus, the unique functions that satisfies this differencial equation with its initial conditions are:
    y=D-\frac{1}{4}-\frac{1}{60}e^{4x}+\frac{1}{15}\cos x-\frac{4}{15}\sin x

    Hope this helps.
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  3. #3
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    Gdansk, Poland
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    Simpler

    Using Laplace transform is simpler. The system is unstable, because it has a pole in the right halfplane (s=4, yes you have mistake in your Laplace transform). Thats why we can see C \exp(4x) in the PerfectHacker's solution.
    Because system is unstable there is no D for which the solution is bounded.
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