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Thread: Solve the linear and separable equation.

  1. #1
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    Solve the linear and separable equation.

    Linear: $\displaystyle x^2y'-x(x^2-2y), y(1) = 5/4
    $

    $\displaystyle x^2y'-x(x^2-2y)= 0$
    $\displaystyle x^2y' = x^3-2xy$

    $\displaystyle y' = \frac{x^3-2xy}{x^2}$

    $\displaystyle y' = x - \frac{2y}{x}$

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

    $\displaystyle e^\int (\frac{2}{x}dx) = ....x^2$

    $\displaystyle x^2y'+x^2(\frac{2}{x})y = x^3$

    $\displaystyle \int(\frac{d}{dx} x^2y) dx = \int x^3 dx$

    $\displaystyle x^2y = \frac{x^4}{4}$

    $\displaystyle y = \frac{x^2}{2}$

    ...lost, don't know what I'm doing wrong.

    Separable: $\displaystyle x^2y'+secy = 0$, $\displaystyle y(1)= \frac{\pi}{6}$

    ?
    Last edited by A Beautiful Mind; Sep 11th 2010 at 02:38 PM.
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  2. #2
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    Quote Originally Posted by A Beautiful Mind View Post
    Linear: $\displaystyle x^2y'-x(x^2-2y), y(1) = 5/4
    $

    $\displaystyle x^2y'-x(x^2-2y)= 0$
    $\displaystyle x^2y' = x^3-2xy$

    $\displaystyle y' = \frac{x^3-2xy}{x^2}$

    $\displaystyle y' = x - \frac{2y}{x}$

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

    $\displaystyle e^\int (\frac{2}{x}dx) = ....x^2$

    $\displaystyle x^2y'+x^2(\frac{2}{x})y = x^3$

    $\displaystyle \int(\frac{d}{dx} x^2y) dx = \int x^3 dx$

    $\displaystyle x^2y = \frac{x^4}{4}$ Mr F adds: + C

    $\displaystyle y = \frac{x^2}{2}$ Mr F says: That 2 should be a 4.

    ...lost, don't know what I'm doing wrong.
    Make the appropriate corrections to get the correct answer.

    Quote Originally Posted by A Beautiful Mind View Post
    Separable: $\displaystyle x^2y'+secy = 0$, $\displaystyle y(1)= \frac{\pi}{6}$

    ?
    $\displaystyle \displaystyle x^2 \frac{dy}{dx} + \frac{1}{\cos y} = 0 \Rightarrow \int \cos y \, dy = \int \frac{1}{x^2} \, dx$.

    And don't make the mistake of forgetting the "+ C" (like you did in the first question) when you do the integrations.
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    Okay, thank you for the tips. I had \frac{x^4}{4} on my paper, but I didn't write it down. And yeah, thanks for the reminder on the plus C.

    I'm not getting where y(1) = 5/4 or (1, 5/4) comes into play here. Not getting a 5/4 when I plug in a 1 for x.
    $\displaystyle
    y = \frac{x^2}{4}+C$
    $\displaystyle
    y - \frac{x^2}{4} = C$
    $\displaystyle \frac{5-1}{4} = C$

    $\displaystyle C = 1$

    $\displaystyle Y(1) = 5/4$

    $\displaystyle 5/4 = \frac{x^4}{4}+1$
    $\displaystyle
    5/4 = \frac{1}{4}+1$
    $\displaystyle
    5/4 = 5/4$
    Last edited by A Beautiful Mind; Sep 11th 2010 at 04:35 PM.
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  4. #4
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    The point is that when you plug in $\displaystyle x = 1$ and $\displaystyle y = \frac{5}{4}$ you should be able to solve for C, which you have done.

    This means the solution to your DE is

    $\displaystyle y = \frac{x^2}{4} + 1$.


    Also for the second, I believe you'll find it's actually

    $\displaystyle \int{\cos{y}\,dy} = -\int{\frac{1}{x^2}\,dx}$.
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  5. #5
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    Yeah, I figured he must've made a typo or something. Solved that one a bit ago. Thanks guys.
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