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Math Help - homogenous differential equation help

  1. #1
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    homogenous differential equation help

    Hi

    I don't understand how solve the homogenous equation

    I set z=y/x therefore dy/dx=(dz/dx)x + z

    however i wasn't able to fully solve it

    i have attached two questions



    thanks
    Attached Thumbnails Attached Thumbnails homogenous differential equation help-homogenous-equation.gif   homogenous differential equation help-homogenous-equation.jpg  
    Last edited by rpatel; May 31st 2009 at 03:45 PM.
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  2. #2
    Super Member Random Variable's Avatar
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     \frac {\mathrm{d}y}{\mathrm{d}x} = \frac {x^{2}-3y^{2}}{xy} = \frac {x}{y} - 3 \frac {y}{x}

    As you said, make the substiution  z = \frac {y}{x}

    then  \frac { \mathrm{d} y}{\mathrm{d}x} = x \frac {\mathrm{d}z}{\mathrm{d}x} + z

    so  x \frac {\mathrm{d}z}{\mathrm{d}x} + z = \frac {1}{z} - 3z = \frac {1-3z^{2}}{z}

     \frac {z}{1-4z^{2}} \ \mathrm{d}z = \frac {1}{x} \ \mathrm{d}x

    Integrate both sides and then substitute for z.
    Last edited by Random Variable; May 31st 2009 at 04:02 PM.
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  3. #3
    Member pberardi's Avatar
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    Quote Originally Posted by rpatel View Post
    Hi

    I don't understand how solve the homogenous equation

    I set z=y/x therefore dy/dx=(dz/dx)x + z

    however i wasn't able to fully solve it

    i have attached two questions



    thanks
    For the second.
    dy/dx = 1/x[y + (x)/sin(y/x)]
    = y/x + [1/sin(y/x)]
    let v = y/x, y = vx, dy/dx = (x)dv/dx + v
    (x)dv/dx + v = v + 1/sin(v)
    (x)dv/dx = 1/sin(v)
    sin(v)dv = 1/x dx
    -cos(v) = ln|x| + C1
    cos(v) = -ln|x| + C1
    v = cos^-1(C1-ln|x|)
    y = xcos^-1(C1-ln|x|)

    If you don't believe me check this out.
    x*y*y' - y(y + x/sin(y/x)) = 0 - Wolfram|Alpha

    Also, here is a link that shows you how to show that a differential equation is homogeneous Homogeneous Differential Equations: Example 2
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  4. #4
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    Hello, rpatel!

    Solve the homogenous equation: . xy\frac{dy}{dx} \:=\:x^2 - 3y^2
    Divide by xy\!:\;\;\frac{dy}{dx} \:=\:\frac{x}{y} - 3\frac{y}{x}

    Let: z = \frac{y}{x} \quad\Rightarrow\quad y =xy \quad\Rightarrow\quad \frac{dy}{dx} \:=\:x\frac{dz}{dx} + z

    Substitute: . x\frac{dz}{dx} + z \:=\:\frac{1}{z} - 3z \:=\:\frac{1-3z^2}{z}

    . . Then: . \frac{dz}{dx} \:=\:\frac{1-3z^2}{z} - z \:=\:\frac{1-4z^2}{z}

    Separate variables: . \frac{z}{1-4z^2}\,dz \:=\:\frac{dx}{x}


    Integrate: . \int\frac{z\,dz}{1-4z^2} \;=\;\int\frac{dx}{x} \quad\Rightarrow\quad -\tfrac{1}{8}\ln\left|1-4z^2\right| \;=\;\ln x + c_1


    Multiply by -8: . \ln|1-4z^2| \:=\:\text{-}8\ln x + c_2 \;=\;\ln(x^{\text{-}8}) + \ln(c_3) \;=\;\ln(c_3x^{\text{-}8})

    Exponentiate: . 1 - 4z^2 \;=\;c_3x^{\text{-}8} \quad\Rightarrow\quad 1 - 4\frac{y^2}{x^2} \;=\;c_3x^{\text{-}8}

    Multiply by x^2\!:\;\;x^2 - 4y^2 \;=\;c_3x^{-6} \quad\Rightarrow\quad \text{-}4y^2 \;=\;\text{-}x^2 + c_3x^{-6}


    Divide by -4: . y^2 \;=\;\tfrac{1}{4}x^2 + C x^{-6}

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