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Math Help - Homogenous D.E

  1. #1
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    Homogenous D.E

    Any help solving homogenous differntial equation would be greatly appreciated:

    dy/dx=(-2x^2+2y^2)/xy

    The following is what i've worked out (could be incorrect):
    dy/dx=2(y/x-x/y)
    let v=y/x, dy/dx=v+x*dv/dx
    v+x*dv/dx=2(v-1/v), x*dv/dx=2(v-1/v)-v
    therefore, x*dv/dx=(v^2-2)/v

    1/x dx=v/(v^2-2) dv
    -ln(x)+C=1/2 ln(v^2-2)
    C/x=1/2(y^2/x^2-2)
    C/x=y^2/2x^2-1
    C/x+1=y^2/2x^2
    y^2=2x^2(C/x+1)
    y=sqrt(2x^2(C/x+1))

    I must of gone wrong somewhere as it says the answer is incorrect, i dont know where though...
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by skirk34 View Post
    Any help solving homogenous differntial equation would be greatly appreciated:

    dy/dx=(-2x^2+2y^2)/xy

    The following is what i've worked out (could be incorrect):
    dy/dx=2(y/x-x/y)
    let v=y/x, dy/dx=v+x*dv/dx
    v+x*dv/dx=2(v-1/v), x*dv/dx=2(v-1/v)-v
    therefore, x*dv/dx=(v^2-2)/v

    1/x dx=v/(v^2-2) dv
    -ln(x)+C=1/2 ln(v^2-2)
    C/x=1/2(y^2/x^2-2)
    C/x=y^2/2x^2-1
    C/x+1=y^2/2x^2
    y^2=2x^2(C/x+1)
    y=sqrt(2x^2(C/x+1))

    I must of gone wrong somewhere as it says the answer is incorrect, i dont know where though...
    \int 1/x dx=\int v/(v^2-2) dv

    \ln(Ax)= \frac{1}{2} \ln(v^2-2)

    A is an arbitary constant, and whatever transformation is applied to it I will use the same symbol for the result, so it is not the same thing in every expression that follows:

    \ln(Ax^2)= \ln(v^2-2)

    so:

    Ax^2=v^2-2

    v^2=Ax^2+2

    y^2=x^2(Ax^2+2)

    etc

    RonL
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  3. #3
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    cool, so it was just a simple mistake i made in evaluating the logarithms. Thxs
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by skirk34 View Post
    cool, so it was just a simple mistake i made in evaluating the logarithms. Thxs
    No look at the sign of the integral of 1/x (Note there is also an implicit assumption that x is positive as otherwise the log should be of A|x|)

    RonL
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