Clairaut's equation: Find the general solution of $x^2 (y-xy') =y(y') ^2$ if the singular solution doesn't...












1














Question:
Find the general solution of



$$x^2 (y-xy') =y(y') ^2$$



if the singular solution doesn't exist.



Now, I know that it has to be solved by Clairaut's equation.



However, the given equation is not of the form $y=px+f(p)$ where $p=y'$ and cannot, as far as I know, be reduced to this form.



I even tried arranging the equation in the form of a quadratic in $y'$ but when I find the solution it yields a rather big equation that doesn't really lead anywhere.



Please let me know if you have any suggestions.






differential-equations







share|cite|improve this question





























    1














    Question:
    Find the general solution of



    $$x^2 (y-xy') =y(y') ^2$$



    if the singular solution doesn't exist.



    Now, I know that it has to be solved by Clairaut's equation.



    However, the given equation is not of the form $y=px+f(p)$ where $p=y'$ and cannot, as far as I know, be reduced to this form.



    I even tried arranging the equation in the form of a quadratic in $y'$ but when I find the solution it yields a rather big equation that doesn't really lead anywhere.



    Please let me know if you have any suggestions.






    differential-equations







    share|cite|improve this question



























      1












      1








      1


      1





      Question:
      Find the general solution of



      $$x^2 (y-xy') =y(y') ^2$$



      if the singular solution doesn't exist.



      Now, I know that it has to be solved by Clairaut's equation.



      However, the given equation is not of the form $y=px+f(p)$ where $p=y'$ and cannot, as far as I know, be reduced to this form.



      I even tried arranging the equation in the form of a quadratic in $y'$ but when I find the solution it yields a rather big equation that doesn't really lead anywhere.



      Please let me know if you have any suggestions.






      differential-equations







      share|cite|improve this question















      Question:
      Find the general solution of



      $$x^2 (y-xy') =y(y') ^2$$



      if the singular solution doesn't exist.



      Now, I know that it has to be solved by Clairaut's equation.



      However, the given equation is not of the form $y=px+f(p)$ where $p=y'$ and cannot, as far as I know, be reduced to this form.



      I even tried arranging the equation in the form of a quadratic in $y'$ but when I find the solution it yields a rather big equation that doesn't really lead anywhere.



      Please let me know if you have any suggestions.






      differential-equations




      differential-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 27 '18 at 2:38









      Shaun

      8,760113680




      8,760113680










      asked Nov 27 '18 at 2:11









      MadCap

      213




      213






















          1 Answer
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          oldest

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          Hint



          Rewrite the equation as
          $$x^2 left(y-frac{x}{x'}right)-frac{y}{x'^2}=0$$ Now, let $x=sqrt{y z}$ which makes
          $$frac{y^2 z left(y^2 z'^2-z^2-4right)}{left(y z'+zright)^2}=0$$ Excluding the trivial solution, we are left with
          $$y^2 z'^2-z^2-4=0$$ that is to say
          $$frac{z'^2}{z^2+4}=frac 1 {y^2}implies frac{z'}{sqrt{z^2+4}}=pm frac 1y$$ which seems to be simple to integrate.






          share|cite|improve this answer





















          • Shouldn't the denominator be yz'+zy' when you substitute for x or am I missing something here?
            – MadCap
            Nov 28 '18 at 0:27










          • @MadCap. Since $y$ became the variable, $y'=1$. I double check and all of this seems to be correct.
            – Claude Leibovici
            Nov 28 '18 at 3:08










          • Oh now it's more clear.
            – MadCap
            Nov 28 '18 at 13:18











          Your Answer





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          1 Answer
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          1 Answer
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          active

          oldest

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          active

          oldest

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          1














          Hint



          Rewrite the equation as
          $$x^2 left(y-frac{x}{x'}right)-frac{y}{x'^2}=0$$ Now, let $x=sqrt{y z}$ which makes
          $$frac{y^2 z left(y^2 z'^2-z^2-4right)}{left(y z'+zright)^2}=0$$ Excluding the trivial solution, we are left with
          $$y^2 z'^2-z^2-4=0$$ that is to say
          $$frac{z'^2}{z^2+4}=frac 1 {y^2}implies frac{z'}{sqrt{z^2+4}}=pm frac 1y$$ which seems to be simple to integrate.






          share|cite|improve this answer





















          • Shouldn't the denominator be yz'+zy' when you substitute for x or am I missing something here?
            – MadCap
            Nov 28 '18 at 0:27










          • @MadCap. Since $y$ became the variable, $y'=1$. I double check and all of this seems to be correct.
            – Claude Leibovici
            Nov 28 '18 at 3:08










          • Oh now it's more clear.
            – MadCap
            Nov 28 '18 at 13:18
















          1














          Hint



          Rewrite the equation as
          $$x^2 left(y-frac{x}{x'}right)-frac{y}{x'^2}=0$$ Now, let $x=sqrt{y z}$ which makes
          $$frac{y^2 z left(y^2 z'^2-z^2-4right)}{left(y z'+zright)^2}=0$$ Excluding the trivial solution, we are left with
          $$y^2 z'^2-z^2-4=0$$ that is to say
          $$frac{z'^2}{z^2+4}=frac 1 {y^2}implies frac{z'}{sqrt{z^2+4}}=pm frac 1y$$ which seems to be simple to integrate.






          share|cite|improve this answer





















          • Shouldn't the denominator be yz'+zy' when you substitute for x or am I missing something here?
            – MadCap
            Nov 28 '18 at 0:27










          • @MadCap. Since $y$ became the variable, $y'=1$. I double check and all of this seems to be correct.
            – Claude Leibovici
            Nov 28 '18 at 3:08










          • Oh now it's more clear.
            – MadCap
            Nov 28 '18 at 13:18














          1












          1








          1






          Hint



          Rewrite the equation as
          $$x^2 left(y-frac{x}{x'}right)-frac{y}{x'^2}=0$$ Now, let $x=sqrt{y z}$ which makes
          $$frac{y^2 z left(y^2 z'^2-z^2-4right)}{left(y z'+zright)^2}=0$$ Excluding the trivial solution, we are left with
          $$y^2 z'^2-z^2-4=0$$ that is to say
          $$frac{z'^2}{z^2+4}=frac 1 {y^2}implies frac{z'}{sqrt{z^2+4}}=pm frac 1y$$ which seems to be simple to integrate.






          share|cite|improve this answer












          Hint



          Rewrite the equation as
          $$x^2 left(y-frac{x}{x'}right)-frac{y}{x'^2}=0$$ Now, let $x=sqrt{y z}$ which makes
          $$frac{y^2 z left(y^2 z'^2-z^2-4right)}{left(y z'+zright)^2}=0$$ Excluding the trivial solution, we are left with
          $$y^2 z'^2-z^2-4=0$$ that is to say
          $$frac{z'^2}{z^2+4}=frac 1 {y^2}implies frac{z'}{sqrt{z^2+4}}=pm frac 1y$$ which seems to be simple to integrate.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 27 '18 at 3:49









          Claude Leibovici

          119k1157132




          119k1157132












          • Shouldn't the denominator be yz'+zy' when you substitute for x or am I missing something here?
            – MadCap
            Nov 28 '18 at 0:27










          • @MadCap. Since $y$ became the variable, $y'=1$. I double check and all of this seems to be correct.
            – Claude Leibovici
            Nov 28 '18 at 3:08










          • Oh now it's more clear.
            – MadCap
            Nov 28 '18 at 13:18


















          • Shouldn't the denominator be yz'+zy' when you substitute for x or am I missing something here?
            – MadCap
            Nov 28 '18 at 0:27










          • @MadCap. Since $y$ became the variable, $y'=1$. I double check and all of this seems to be correct.
            – Claude Leibovici
            Nov 28 '18 at 3:08










          • Oh now it's more clear.
            – MadCap
            Nov 28 '18 at 13:18
















          Shouldn't the denominator be yz'+zy' when you substitute for x or am I missing something here?
          – MadCap
          Nov 28 '18 at 0:27




          Shouldn't the denominator be yz'+zy' when you substitute for x or am I missing something here?
          – MadCap
          Nov 28 '18 at 0:27












          @MadCap. Since $y$ became the variable, $y'=1$. I double check and all of this seems to be correct.
          – Claude Leibovici
          Nov 28 '18 at 3:08




          @MadCap. Since $y$ became the variable, $y'=1$. I double check and all of this seems to be correct.
          – Claude Leibovici
          Nov 28 '18 at 3:08












          Oh now it's more clear.
          – MadCap
          Nov 28 '18 at 13:18




          Oh now it's more clear.
          – MadCap
          Nov 28 '18 at 13:18


















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