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Math Help - First order linear equation

  1. #1
    Super Member Deadstar's Avatar
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    First order linear equation

    Where am I going wrong here?!?

    Find the general solution of the following equation;

    x \frac{dy}{dx} + 2y = x^3

    i.e. \frac{dy}{dx} + \frac{2}{x}y = x^2

    So, my working.

    Using the whole (ry)' = ry' + r'y = r(y' + py) = rf thing where p = 2/x and f = x^2...

    r' = rp = \frac{2}{x}r

    => (log(r))' = \frac{2}{x}

    => log(r) = log(\frac{2}{x})

    So r = \frac{2}{x}

    => (\frac{2y}{x})' = 2x

    => \frac{2y}{x} = x^2 + C

    => y = \frac{x^3}{2} + \frac{Cx}{2}

    The answer given in the tutorial is (no working given) y = \frac{x^3}{5} + Cx^{-2} and I have no idea how to get that though I figure the problem lies with the log bit... Any help?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello,
    Quote Originally Posted by Deadstar View Post
    Where am I going wrong here?!?

    Find the general solution of the following equation;

    x \frac{dy}{dx} + 2y = x^3

    (...)

    The answer given in the tutorial is (no working given) y = \frac{x^3}{5} + Cx^{-2} and I have no idea how to get that though I figure the problem lies with the log bit... Any help?
    Instead of dividing by x try multiplying by x :

    x \frac{\mathrm{d}y}{\mathrm{d}x} + 2y = x^3\implies <br />
x^2 \frac{\mathrm{d}y}{\mathrm{d}x} + 2xy = x^4

    Now note that the LHS of the last equality is \frac{\mathrm{d}}{\mathrm{d}x}(x^2y) and you are done.
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  3. #3
    Super Member Deadstar's Avatar
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    How clever.

    Only thing is, i have a resit in methods of applied maths in a few days and one of the things i did wrong on the first exam was that thing where you use the (ry)' = rf thing and i know thats what you've sort of done here but i need to know where i go wrong in my working above.
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Deadstar View Post
    i need to know where i go wrong in my working above.
    Quote Originally Posted by Deadstar View Post
    => (log(r))' = \frac{2}{x}

    => log(r) = log(\frac{2}{x})
    This step is wrong. Integrating both sides of \left(\log r\right)' = \frac{2}{x} you should get  \log r = 2\log x+C=\log x^2+C hence r=\exp\left(\log x^2+C \right)=x^2 \underbrace{\exp C}_{\text{constant}}=K x^2. Using (ry)'=rf you'll get the result you're expecting.
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