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Thread: Find the solution in the form of a quadratic function of t, where a,b, and c tbd

  1. #1
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    Find the solution in the form of a quadratic function of t, where a,b, and c tbd

    Find a solution x=x(t) of the equation x'+2x=t2+4t+7 in the form of a quadratic function of t, that is, of the form x(t) = at2+bt +c, where a,b, and c are to be determined.

    At first I thought you would solve this by finding the integrating factor, but after speaking with the teacher he said to plug in the form x(t)=at2+bt+c into the Diff. Eq. and use the resulting expression to solve for a,b, and c by comparing the expression on the left side to the expression on the right side.

    So now I get this, but I am not sure what to do from here.

    x'(t)=2t+4
    2t2+10t+18 = t2+4t+7
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  2. #2
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    Re: Find the solution in the form of a quadratic function of t, where a,b, and c tbd

    $x(t)=at^2+bt+c $
    $x'(t)=2at+b $

    Plugging in:

    $2at+b+2at^2+2bt+2c=t^2+4t+7$

    Equate terms:
    $2at^2=t^2$
    $2 (a+b)t=4t $
    $b+2c=7$

    This gives:
    $a=\dfrac {1}{2},b=\dfrac {3}{2}, c=\dfrac {11}{4} $
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  3. #3
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    Re: Find the solution in the form of a quadratic function of t, where a,b, and c tbd

    thank you.
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