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Thread: 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;

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

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

    So, my working.

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

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

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

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

    So $\displaystyle r = \frac{2}{x}$

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

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

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

    The answer given in the tutorial is (no working given) $\displaystyle y = \frac{x^3}{5} + Cx^{-2}$ 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;

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

    (...)

    The answer given in the tutorial is (no working given) $\displaystyle y = \frac{x^3}{5} + Cx^{-2}$ 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 $\displaystyle x$ try multiplying by $\displaystyle x$ :

    $\displaystyle x \frac{\mathrm{d}y}{\mathrm{d}x} + 2y = x^3\implies
    x^2 \frac{\mathrm{d}y}{\mathrm{d}x} + 2xy = x^4$

    Now note that the LHS of the last equality is $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}(x^2y)$ 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
    => $\displaystyle (log(r))' = \frac{2}{x} $

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