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Math Help - Solve the linear and separable equation.

  1. #1
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    Solve the linear and separable equation.

    Linear:  x^2y'-x(x^2-2y), y(1) = 5/4<br />

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

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

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

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

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

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

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

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

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

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

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

    ?
    Last edited by A Beautiful Mind; September 11th 2010 at 03:38 PM.
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  2. #2
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    Quote Originally Posted by A Beautiful Mind View Post
    Linear:  x^2y'-x(x^2-2y), y(1) = 5/4<br />

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

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

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

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

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

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

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

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

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

    ?
    \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.
    <br />
y = \frac{x^2}{4}+C
    <br />
y - \frac{x^2}{4} = C
    \frac{5-1}{4} = C

    C = 1

    Y(1) = 5/4

    5/4 = \frac{x^4}{4}+1
    <br />
5/4 = \frac{1}{4}+1
    <br />
5/4 = 5/4
    Last edited by A Beautiful Mind; September 11th 2010 at 05:35 PM.
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  4. #4
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    The point is that when you plug in x = 1 and y = \frac{5}{4} you should be able to solve for C, which you have done.

    This means the solution to your DE is

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


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

    \int{\cos{y}\,dy} = -\int{\frac{1}{x^2}\,dx}.
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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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