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Math Help - ode's and constants

  1. #1
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    Question ode's and constants

    i want to find the values of the real constants a,b,c for which x(t) = ae^(bt) +ct is a solution to the ODE d^2/dt^2 = 16e^(4t) / xThis is what i have done so far

    i differentiated x to get abe^(bt) + c
    then i differentialiated that again to get (ab^2)*e^(bt)

    then i substituted x and x" into the ODE given to get

    (ab^2)*e^(bt) = 16e^(4t) / ae^(bt) +ct

    this is where i got stuck..am i on the right path..from this part i need to find a b and c..any help thankz
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  2. #2
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    i just worked out that a=1 b=4 and c=0
    is that right ..did anyone else get that?
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  3. #3
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    correction of my answer

    Quote Originally Posted by dopi View Post
    i just worked out that a=1 b=4 and c=0
    is that right ..did anyone else get that?
    acutally..i just went through it again..and it didnt work out...i tried new values for a,b,c

    a=2 b=2 and c=0...........did anyone else get that???
    thankz
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  4. #4
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    Quote Originally Posted by dopi View Post
    i want to find the values of the real constants a,b,c for which x(t) = ae^(bt) +ct is a solution to the ODE d^2/dt^2 = 16e^(4t) / xThis is what i have done so far
    The differencial equation is,

    x''=\frac{16 e^{-4t}}{x}
    x\cdot x''=16e^{-4t}.
    Now we are told that,
    x=ae^{bt}+ct
    Is some solution.
    Then by substituting it into the differencial equation we have equality.
    First we find the first derivative,
    x'=abe^{bt}+c
    x''=ab^2e^{bt}
    Thus,
    (ae^{bt}+ct)(ab^2e^{bt})=16e^{-4t}
    Multiply through,
    a^2b^2e^{2bt}+ab^2cte^{bt}=16e^{-4t}
    Note the left hand side has some multiple of tb^{bt} while the right does not. Thus to match we need it to be zero. Thus, one of the constants (at least) is zero, that is a \mbox{ or } b \mbox{ or } c=0.
    Thus it dies,
    a^2b^2e^{2bt}=16e^{-4t}.
    We see that it cannot be a,b because they must be non-zero in this equation. Thus, c=0.
    The exponents need to match, for equality, that is,
    2b=-4
    b=-2.
    Then by substituting for this value we have,
    a^2 (-2)^2 e^{-4t}=16e^{-4t}
    We immediately see that a=\pm 2.

    Thus, there are two possibilities,
    (a,b,c)=(-2,2,0)
    (a,b,c)=(2,2,0)
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  5. #5
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    sorry can u explain again why one of the constants has to be zero? i dont quite understand that

    thanks
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  6. #6
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    Smile

    [QUOTE=ThePerfectHacker;38412]The differencial equation is,

    x''=\frac{16 e^{-4t}}{x}

    you made a mistake on e^(-4t), its actually a postive 4, but thats alright, still came out with the same answer, thanks
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  7. #7
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    Quote Originally Posted by chogo View Post
    sorry can u explain again why one of the constants has to be zero? i dont quite understand that

    thanks
    It is basicaly the same thing you do when you do the Method of Undetermined coefficients, I assume you heard of it. If note you assume a solution has a certain form like here and substitute it and solve for the constants.
    For example,
    2\sin t +t\cos t=A\sin t +2Bt\cos t +Ce^t
    For the right and left hand sides to be equal you make the coefficients equal to each other.

    1)There is a 2\sin t on left side and there is A\sin t on right side. To be equal the coefficient must match, that is, A=2.

    2)There is t\cos t on left side and there is 2B\cos t on right side. To be equal the coefficient must match, that is, 2B=1. Thus, B=1/2.

    3)And there is not e^t on left hand side. Thus, C=0
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  8. #8
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    yes thanks alot never heard of Method of Undetermined coefficients. But read up on it, thanks alot it was very interesting.
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