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Math Help - Basic Differential Equations

  1. #1
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    Basic Differential Equations

    We have just had an introduction to differential equations, and I need a little help as I left a bit confused and am still fumbling around with two problems:

    Suppose Q=Ce^{-kt} and that \frac {dQ} {dt} =  -0.06Q

    What does this say about k and C?

    At this point, I thought that  k = -.06 and that C could be any value at all, but there are apparently two different answers and I am not sure if what I had mentioned before is even correct.


    The other one that I am having troubles with is:

    <br />
x\frac{dy} {dx} - 4y = 0

    I found that the general solution was Cx^4, but the problem asks for a second solution y=Cx^n that might not be a general solution and which may have a different value of n than the one that was previously found. Could the answer be y=0 or is there an other answer that I am just not thinking of?

    Thanks for the help everyone.
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  2. #2
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    Quote Originally Posted by Spudwad View Post
    Suppose Q=Ce^{-kt} and that \frac {dQ} {dt} = -0.06Q

    What does this say about k and C?

    At this point, I thought that  k = -.06 and that C could be any value at all, but there are apparently two different answers and I am not sure if what I had mentioned before is even correct.
    Solve the differential equation. It is separable:

    \frac{dQ}{0.06Q} = -dt

    \int \frac{dQ}{Q} = \int -0.06 dt

    \ln Q = -0.06t + C_1

    Q = e^{-0.06t + C_1}

    Q = e^{C_1}e^{-0.06t}

    Q = Ce^{-0.06t}

    Unless there is additional information, it seems that k = -0.06 and C could be any positive constant.
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  3. #3
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    For the second, divide though by x, then multiply through by a suitable integrating factor... 1/x^4.

    Edit: please excuse earlier typo. Also, just in case a picture helps...



    ... where



    ... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x.

    So you now have an exact product-rule derivative on the left-had side, and you might be able to fill in the blanks. Then check by seeing, with or without using the pic, that differentiating gives you the bottom row.

    As with the first problem, you haven't included any info that would enable you to then establish the value of c, the constant of integration. (But you perhaps aren't meant to.)

    But how were you meant to choose 1/x^4 as the integrating factor? Basically, by making it e to the power of an anti-derivative of -4/x.

    -4/x being what you have next to y after dividing through by x in the first place.

    So how were you meant to know to divide through by x? See Integrating Factor
    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; March 18th 2010 at 03:21 AM. Reason: typo and pics
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