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Math Help - Nonlinear differential equations of the second order – need your help

  1. #1
    Newbie Metrika's Avatar
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    Question Nonlinear differential equations of the second order – need your help

    Please help me solve these two nonlinear differential equations of the second order

    1) x^3y''=(y-xy')(y-xy'-x)

    2) x^2(yy''-(y')^2)+xyy'=(2xy'-3y)\sqrt{x^3}


    I do not even know how to begin
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Metrika View Post
    Please help me solve these two nonlinear differential equations of the second order

    1) x^3y''=(y-xy')(y-xy'-x)
    \displaystyle{x^3y''=(y-xy')(y-xy'-x)~\Rightarrow~x^3y''=(xy'-y)^2+x(xy'-y)~\Rightarrow}

    \displaystyle{\Rightarrow~y''=x\!\left(\frac{xy'-y}{x^2}\right)^2+\frac{xy'-y}{x^2}~\Rightarrow~y''=x\!\left(\frac{y}{x}\right  )'^2+\left(\frac{y}{x}\right)'\Rightarrow}

    \displaystyle{\Rightarrow~\!\left\{\begin{gathered  }y=xz\hfill\\y'=z+xz'\hfill\\y''=2z'+xz''\hfill\\\  end{gathered}\right\}\!~\Rightarrow~xz''-xz'^2+z'=0}

    Consider two cases:

    1) z'=0~\Rightarrow~z=C~\Rightarrow~y=Cx;

    2) z'\ne0 :

    \displaystyle{\frac{z''}{z'}-z'+\frac{1}{x}=0~\Rightarrow~\left(\ln{z'}-z+\ln{x}\right)^\prime}=0~\Rightarrow~\ln|z'|-z+\ln|x|=C_1~\Rightarrow}

    \displaystyle{\Rightarrow~\ln|xz'|=C_1+z~\Rightarr  ow~xz'=e^{C_1+z}~\Rightarrow~\int{e^{-(C_1+z)}\,dz}=\int\frac{dx}{x}~\Rightarrow}

    \displaystyle{\Rightarrow~-e^{-(C_1+z)}=\ln|x|+C_2~\Rightarrow~e^{-(C_1+z)}=\ln\left|\frac{1}{C_2x}\right|\Rightarrow  ~-(C_1+z)=\ln\ln\left|\frac{1}{C_2x}\right|\Rightarr  ow}

    \displaystyle{\Rightarrow~z=-C_1-\ln\ln\left|\frac{1}{C_2x}\right|=\ln\frac{C_3}{\l  n|C_2x|}~\Rightarrow~y=x\ln\frac{C_3}{\ln|C_2x|}}
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by Metrika View Post
    Please help me solve these two nonlinear differential equations of the second order

    1) x^3y''=(y-xy')(y-xy'-x)
    substitute t=y-xy' and you'll end up with a homogeneous DE, then put t=ux and the reamining equation will be linear, everything straightforward and you'll get the result.
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  4. #4
    MHF Contributor
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    For the second, divide by xy^2 and then re-write as

    x \dfrac{d}{dx} \left( \dfrac{y'}{y}\right) + \dfrac{y'}{y} = 2 \left( \dfrac{x^{3/2}}{y}\right)^2 \dfrac{d}{dx} \left(\dfrac{y}{x^{3/2}} \right)
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