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Math Help - Differential equation Problem

  1. #1
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    Cool Differential equation Problem

    find the solution of the differential equation ( y2- xy) dx + x2 dy = 0

    Please help me with this question. Thank you
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  2. #2
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    Re: Differential equation Problem

    Hello, Deci!

    \text{Solve: }\:(y^2- xy)\:\!dx + x^2\:\!dy \:=\: 0

    We have: . \frac{dy}{dx} \:=\:\frac{xy-y^2}{x^2} \quad\Rightarrow\quad \frac{dy}{dx} \:=\:\frac{y}{x} - \left(\frac{y}{x}\right)^2

    Let v \,=\,\frac{y}{x}\quad\Rightarrow\quad y \:=\:vx\quad\Rightarrow\quad\frac{dy}{dx} \:=\:v + x\frac{dv}{dx}

    Substitute: . v + x\frac{dv}{dx} \:=\:v - v^2 \quad\Rightarrow\quad x\frac{dv}{dx} \:=\:-v^2 \quad\Rightarrow\quad v^{-2}dv \:=\:-\frac{dx}{x}

    Integrate: . \int v^{-2}dv \:=\:-\int\frac{dx}{x} \quad\Rightarrow\quad -v^{-1} \:=\:-\ln x + c

    . . . \frac{1}{v} \:=\:\ln x + c \:=\:\ln x + \ln C \:=\:\ln(Cx)

    . . . v \:=\:\frac{1}{\ln(Cx)} \quad\Rightarrow\quad \frac{y}{x} \:=\:\frac{1}{\ln(Cx)}

    Therefore: . y \;=\;\frac{x}{\ln(Cx)}
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  3. #3
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    Re: Differential equation Problem

    Thanks for your help. I have one simple question



    <<<< isn't this supposed to be ln x - c or the plus minus sign in c doesn't matter?

    and the answer in the ans key is ln x + x/y = C, but when i convert using v = 1 / (ln x + C), then y = vx, so y = x /(ln x + c) >>> ln x - x/y = C , why is it minus?

    And is this what ppl call homogenous equation?
    Last edited by Deci; January 23rd 2014 at 06:34 AM.
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  4. #4
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    Re: Differential equation Problem

    Remember that C is an arbitrary constant of integration. In differential equations, you work with these arbitrary constants a lot, so instead of using different letters like:

    y=x+C
    substitute D=-C, so C=-D
    y=x-D

    it's typical to just "recycle" the same letter C:

    y=x+C
    replace C with -C
    y=x-C

    Since C is an arbitrary constant, these two equations are equivalent.

    There is a difference between Soroban's answer and the answer in the book, though: ln x + x/y = C is not the same as ln x - x/y = C, no matter how much you fiddle with C. Soroban's answer is correct.

    Homogeneous is used for two different things in differential equations - check the Wikipedia article.

    - Hollywood
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  5. #5
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    Re: Differential equation Problem

    Okay thanks for clearing things up hollywood
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