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Math Help - Homogeneous polar equation substitutions

  1. #1
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    Homogeneous polar equation substitutions

    Hello all,

    I've spent the better part of an hour on a single problem. I've attempted it several ways but have yet to find the right answer - every attempt is resulting in a different answer.

    The method I'm supposed to be using is substitution. The problem is (2x^2-3xy)y'=x^2+2xy-3y^2

    The correct answer is y(x)=x*|ln(y^2/x)+C|.

    My latest answer is y=2*x*ln(y/x) - x*ln|x| as a result of integrating (2-v)dv = -(1/x)dx where v=y/x. My previous two attempts had vastly different answers. I have a sneaking suspicion my problem is in my algebra before I start integrating.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Homogeneous polar equation substitutions

    We are given to solve:

    \left(2x^2-3xy \right)y'=x^2+2xy-3y^2

    \frac{dy}{dx}=\frac{x^2+2xy-3y^2}{2x^2-3xy}=\frac{1+2\left(\frac{y}{x} \right)-3\left(\frac{y}{x} \right)^2}{2-3\left(\frac{y}{x} \right)}

    Now, if we use the substitution:

    v=\frac{y}{x}\implies y=vx\implies \frac{dy}{dx}=v+x\frac{dv}{dx}

    We may write:

    v+x\frac{dv}{dx}=\frac{1+2v-3v^2}{2-3v}

    x\frac{dv}{dx}=\frac{1+2v-3v^2}{2-3v}-v=\frac{1+2v-3v^2-v(2-3v)}{2-3v}=\frac{1}{2-3v}

    \int 2-3v\,dv=\int\frac{1}{x}\,dx

    2v-\frac{3}{2}v^2=\ln|x|+C

    Now, back-substitute for v, however, I don't see how the given result is correct.
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  3. #3
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    Re: Homogeneous polar equation substitutions

    I miss-scribed from my book - that's actually the problem right next to the problem I'm having the problem with! (2x^2-xy)y'=xy-y^2 is the problematic problem.

    I believe my problem to be in the algebra - I'm rusty from not doing anything algebraically tricky for a few years.

    In the problem I gave you, your second line, to the right of the second equal sign, you smack the problem upside the head with some algebra - you multiply through with 1/x^2, correct?

    In my problem it appears I should do the following (Pardon my lack of Latex):

    y'=(x^3+3y^3)/(2x^2-xy) * (1/x^3)/(1/x^3)
    = (1+3(y^3/x^3))/((2/x)-((y/x)*(1/x))

    v=y/x
    y=vx
    dy/dx=v+x(dv/dx)

    v+x(dv/dx)=(1+3v^2)/((2/x)-(v/x) = (x^2*(3v^3+1))/(2x-v)

    At about this point, I think I'm lost again. I expanded what I had and spent half a page algebraically manipulating it.

    This is what I boiled it down to, likely incorrectly: 2/x-v/(x^2) + (dv/dx)*(2/v - 1) = 3v^2+1/v.
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  4. #4
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    Re: Homogeneous polar equation substitutions

    The first thing I notice about (2x^2- xy)y'= xy- y^2 is that all terms are of second degree- x^2, y^2, or xy. That suggests the substitution u= \frac{y}{x} or y= xu. Then y'= xu'+ u and the equation becomes (2x^2- x^2u)(xu'+ u)= x^2u+ x^2u^2. Dividing through by x^2 (2- u)(xu'+ u)= (-xu+ 2x)u'+ 2u- u^2= u- u^2 or x(2- u)u'= -u, which is separable: \frac{(2- u)du}{-u}= (1- \frac{2}{u})du= \frac{dx}{x}.
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