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Math Help - Differential equations

  1. #1
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    Differential equations

    Happy new year to all
    I 've just studying differential equations and I 've some equations need solve, please help me!

    1. dy/dx = (x^2 - y^2) / 2xy
    2. x * dy/dx -y = x^3
    3. (x+ (e^-x)* siny)dx - (y+ (e^-x) cosx)dy = 0
    And which method used to solve these type!
    Thanks!
    Sorry, I don't know how to type math equations in this forum, could anyone show me
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  2. #2
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    Quote Originally Posted by Kamikazi View Post
    Happy new year to all
    I 've just studying differential equations and I 've some equations need solve, please help me!

    1. dy/dx = (x^2 - y^2) / 2xy
    2. x * dy/dx -y = x^3
    3. (x+ (e^-x)* siny)dx - (y+ (e^-x) cosx)dy = 0
    And which method used to solve these type!
    Thanks!
    Sorry, I don't know how to type math equations in this forum, could anyone show me
    1. \frac{dy}{dx} = \frac{x^2 - y^2}{2xy} = \frac{1 - \left(\frac{y}{x}\right)^2}{2 \left(\frac{y}{x}\right)}

    The usual technique is to now make the substitution y = xv.


    2. Re-write as \frac{dy}{dx} - \frac{1}{x} y = x^2. Now use the integrating factor method.

    Note that the integrating factor is e^{\int -\frac{1}{x} \, dx} = e^{- \ln |x|} = e^{\ln \frac{1}{|x|}} = \frac{1}{x}.


    3. I don't have time now to check right now but you might need to find an integrating factor to make it exact.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    1. \frac{dy}{dx} = \frac{x^2 - y^2}{2xy} = \frac{1 - \left(\frac{y}{x}\right)^2}{2 \left(\frac{y}{x}\right)}

    The usual technique is to now make the substitution y = xv.


    2. Re-write as \frac{dy}{dx} - \frac{1}{x} y = x^2. Now use the integrating factor method.

    Note that the integrating factor is e^{\int -\frac{1}{x} \, dx} = e^{- \ln |x|} = e^{\ln \frac{1}{|x|}} = \frac{1}{x}.


    3. I don't have time now to check right now but you might need to find an integrating factor to make it exact.
    Thanks a lot, now I understand
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  4. #4
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    As for the third, it's exact as it stands if the cos x is a cos y.
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