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Math Help - general solution of this differential equation?

  1. #1
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    general solution of this differential equation?

    what's the general solution?

    x*y*y' = y^2 + x*sqrt(4*x^2 + y^2)
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  2. #2
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    Quote Originally Posted by fryfry View Post
    what's the general solution?

    x*y*y' = y^2 + x*sqrt(4*x^2 + y^2)
    Note that if you divide both sides by x^2 the DE can be re-written as


    \displaystyle \frac{y}{x} \frac{dy}{dx} = \left( \frac{y}{x} \right)^2 + \sqrt{4 + \left( \frac{y}{x} \right)^2}.
    Last edited by mr fantastic; September 10th 2010 at 04:18 AM. Reason: Forgot the = sign.
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  3. #3
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    I'm not sure I quite agree. I get

    \displaystyle{\frac{d}{dx}\left(\frac{y^{2}}{x^{2}  }\right)=\frac{2x^{2}yy'-2xy^{2}}{x^{4}}=\frac{2xyy'-2y^{2}}{x^{3}}=\frac{2}{x}\left(\frac{xyy'-y^{2}}{x^{2}}\right)=\frac{2}{x}\sqrt{4+\left(\fra  c{y}{x}\right)^{2}}.}

    That last equality comes from the DE, from arranging the LHS to look like what's in the parentheses. At this point, you could make the substitution u(x)=y(x)/x.

    I was thinking quotient rule for derivatives, but was getting hung up on not having the x squared term in the denominator, which you have to have in order to make the coefficients the same in the numerator.
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  4. #4
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    Quote Originally Posted by Ackbeet View Post
    I'm not sure I quite agree. I get

    \displaystyle{\frac{d}{dx}\left(\frac{y^{2}}{x^{2}  }\right)=\frac{2x^{2}yy'-2xy^{2}}{x^{4}}=\frac{2xyy'-2y^{2}}{x^{3}}=\frac{2}{x}\left(\frac{xyy'-y^{2}}{x^{2}}\right)=\frac{2}{x}\sqrt{4+\left(\fra  c{y}{x}\right)^{2}}.}

    That last equality comes from the DE, from arranging the LHS to look like what's in the parentheses. At this point, you could make the substitution u(x)=y(x)/x.

    I was thinking quotient rule for derivatives, but was getting hung up on not having the x squared term in the denominator, which you have to have in order to make the coefficients the same in the numerator.
    But I'm not differentiating the y^2/x^2, all I did was divide both sides by x^2. The DE in post #2 is correct (well, it is now that I put the = sign in), and the next step is to substitute y = xv.
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  5. #5
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    Fair enough. I think the previous lack of equals sign was throwing me off.
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  6. #6
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    xyy'=y^2+x\sqrt{4*x^2 + y^2}

    May be this way is good
    (1) u^2=4x^2+y^2
    uu'=4x+yy'

    The initial equation becomes
    u'-\frac{u}{x}=1
    \frac{u}{x}=ln|x|+C
    From (1) get y.
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