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Math Help - Urgent: Differential equation.

  1. #1
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    Exclamation Urgent: Differential equation.

    Can anyone give me any pointers with this differential equation.

    I seriously have no clue what I am doing with it (More detail the better )

    Find the particular solution of:
    x^3 dy/dx = y^3 - 3yx^2 , y(1) = 1


    ..... remember, I am really unsure what I am doing with this.

    Thankyou. math_Newbie.
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  2. #2
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    change of variables

    Define u(x)=x^3y(x). Then \frac{du}{dx}=x^3\frac{dy}{dx}+3x^2y(x)
    Substituting in our original differential equation:
    \frac{du}{dx}=y^3-3x^2y+3x^2y
    \frac{du}{dx}=y^3
    \frac{du}{dx}=\left(\frac{u(x)}{x^3}\right)^3
    \frac{du}{dx}=\frac{u^3}{x^9}
    Which is a separable equation; after solving for u(x) (remember the constant of integration), you can use y(x)=\frac{u(x)}{x^3} to obtain y(x), and then use y(1)=1 to find the value of the constant.

    --Kevin C.
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  3. #3
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    Quote Originally Posted by math_Newbie View Post
    Can anyone give me any pointers with this differential equation.

    I seriously have no clue what I am doing with it (More detail the better )

    Find the particular solution of:
    x^3 dy/dx = y^3 - 3yx^2 , y(1) = 1


    ..... remember, I am really unsure what I am doing with this.

    Thankyou. math_Newbie.
    Alternatively: The DE can be written as \frac{dy}{dx} = \left( \frac{y}{x} \right)^3 - 3 \left( \frac{y}{x} \right).

    For DE's of the form \frac{dy}{dx} = f\left( \frac{y}{x}\right) there's a standard substitution that converts it to first order linear:

    v = \frac{y}{x} \Rightarrow y = xv \, and \, \frac{dy}{dx} = v + \frac{dv}{dx}.

    If you make this substitution in your case, you get v + \frac{dv}{dx} = v^3 - 3v \Rightarrow \frac{dv}{dx} = v^3 - 4v which is seperable:


    \int dx = \int \frac{dv}{v^3 - 4v}.


    Since \frac{1}{v^3 - 4v} = \frac{1}{v(v^2 - 4)} = \frac{1}{v(v - 2)(v + 2)}, you have a case for partial fraction decomposition.

    Alternatively, and just to stop Krizalid getting an ulcer, in this one instance I will point out that instead of using partial fractions, you might note that

    \frac{1}{v(v^2 - 4)} = \frac{1}{4} \left( \frac{v^2 - (v^2 - 4)}{v(v^2 - 4)} \right) = \frac{1}{4} \left( \frac{v}{v^2 - 4} - \frac{1}{v}\right) and "it follows easily"

    Either way, once you've solved for v, you've got y: y = xv.
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  4. #4
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    Krizalid's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Alternatively, and just to stop Krizalid getting an ulcer, in this one instance I will point out that instead of using partial fractions
    Amén!

    (Just by sayin' that I recently got an ulcer.)
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