Solve $frac{dx}{dy}+frac{x}{sqrt{x^2+y^2}}=y$












2












$begingroup$


Solve the differential equation



Solve $$frac{dx}{dy}+frac{x}{sqrt{x^2+y^2}}=y$$



My try:



I used $x=y tan z$



$$frac{dx}{dy}=tan z+sec^2 zfrac{dz}{dy}$$



So we get:



$$tan z+sec^2 zfrac{dz}{dy}+sin z=y$$



Any clue from here?






ordinary-differential-equations derivatives trigonometry







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$endgroup$

















    2












    $begingroup$


    Solve the differential equation



    Solve $$frac{dx}{dy}+frac{x}{sqrt{x^2+y^2}}=y$$



    My try:



    I used $x=y tan z$



    $$frac{dx}{dy}=tan z+sec^2 zfrac{dz}{dy}$$



    So we get:



    $$tan z+sec^2 zfrac{dz}{dy}+sin z=y$$



    Any clue from here?






    ordinary-differential-equations derivatives trigonometry







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      2



      $begingroup$


      Solve the differential equation



      Solve $$frac{dx}{dy}+frac{x}{sqrt{x^2+y^2}}=y$$



      My try:



      I used $x=y tan z$



      $$frac{dx}{dy}=tan z+sec^2 zfrac{dz}{dy}$$



      So we get:



      $$tan z+sec^2 zfrac{dz}{dy}+sin z=y$$



      Any clue from here?






      ordinary-differential-equations derivatives trigonometry







      share|cite|improve this question









      $endgroup$




      Solve the differential equation



      Solve $$frac{dx}{dy}+frac{x}{sqrt{x^2+y^2}}=y$$



      My try:



      I used $x=y tan z$



      $$frac{dx}{dy}=tan z+sec^2 zfrac{dz}{dy}$$



      So we get:



      $$tan z+sec^2 zfrac{dz}{dy}+sin z=y$$



      Any clue from here?






      ordinary-differential-equations derivatives trigonometry




      ordinary-differential-equations derivatives trigonometry






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 28 at 13:09









      Umesh shankarUmesh shankar

      3,07931220




      3,07931220






















          1 Answer
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          2












          $begingroup$

          Hint:



          Let $x=yu$ ,



          Then $dfrac{dx}{dy}=ydfrac{du}{dy}+u$



          $therefore ydfrac{du}{dy}+u+dfrac{yu}{sqrt{y^2u^2+y^2}}=y$



          $ydfrac{du}{dy}+u+dfrac{u}{sqrt{u^2+1}}=y$



          $ydfrac{du}{dy}=y-u-dfrac{u}{sqrt{u^2+1}}$



          $left(y-u-dfrac{u}{sqrt{u^2+1}}right)dfrac{dy}{du}=y$



          This belongs to an Abel equation of the second kind.



          Let $v=y-u-dfrac{u}{sqrt{u^2+1}}$ ,



          Then $y=v+u+dfrac{u}{sqrt{u^2+1}}$



          $dfrac{dy}{du}=dfrac{dv}{du}+1+dfrac{1}{(u^2+1)^frac{3}{2}}$



          $therefore vleft(dfrac{dv}{du}+1+dfrac{1}{(u^2+1)^frac{3}{2}}right)=v+u+dfrac{u}{sqrt{u^2+1}}$



          $vdfrac{dv}{du}+left(1+dfrac{1}{(u^2+1)^frac{3}{2}}right)v=v+u+dfrac{u}{sqrt{u^2+1}}$



          $vdfrac{dv}{du}=-dfrac{v}{(u^2+1)^frac{3}{2}}+u+dfrac{u}{sqrt{u^2+1}}$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Very nice but still i have no clue to proceed from the last step.
            $endgroup$
            – Umesh shankar
            Jan 31 at 14:45












          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Hint:



          Let $x=yu$ ,



          Then $dfrac{dx}{dy}=ydfrac{du}{dy}+u$



          $therefore ydfrac{du}{dy}+u+dfrac{yu}{sqrt{y^2u^2+y^2}}=y$



          $ydfrac{du}{dy}+u+dfrac{u}{sqrt{u^2+1}}=y$



          $ydfrac{du}{dy}=y-u-dfrac{u}{sqrt{u^2+1}}$



          $left(y-u-dfrac{u}{sqrt{u^2+1}}right)dfrac{dy}{du}=y$



          This belongs to an Abel equation of the second kind.



          Let $v=y-u-dfrac{u}{sqrt{u^2+1}}$ ,



          Then $y=v+u+dfrac{u}{sqrt{u^2+1}}$



          $dfrac{dy}{du}=dfrac{dv}{du}+1+dfrac{1}{(u^2+1)^frac{3}{2}}$



          $therefore vleft(dfrac{dv}{du}+1+dfrac{1}{(u^2+1)^frac{3}{2}}right)=v+u+dfrac{u}{sqrt{u^2+1}}$



          $vdfrac{dv}{du}+left(1+dfrac{1}{(u^2+1)^frac{3}{2}}right)v=v+u+dfrac{u}{sqrt{u^2+1}}$



          $vdfrac{dv}{du}=-dfrac{v}{(u^2+1)^frac{3}{2}}+u+dfrac{u}{sqrt{u^2+1}}$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Very nice but still i have no clue to proceed from the last step.
            $endgroup$
            – Umesh shankar
            Jan 31 at 14:45
















          2












          $begingroup$

          Hint:



          Let $x=yu$ ,



          Then $dfrac{dx}{dy}=ydfrac{du}{dy}+u$



          $therefore ydfrac{du}{dy}+u+dfrac{yu}{sqrt{y^2u^2+y^2}}=y$



          $ydfrac{du}{dy}+u+dfrac{u}{sqrt{u^2+1}}=y$



          $ydfrac{du}{dy}=y-u-dfrac{u}{sqrt{u^2+1}}$



          $left(y-u-dfrac{u}{sqrt{u^2+1}}right)dfrac{dy}{du}=y$



          This belongs to an Abel equation of the second kind.



          Let $v=y-u-dfrac{u}{sqrt{u^2+1}}$ ,



          Then $y=v+u+dfrac{u}{sqrt{u^2+1}}$



          $dfrac{dy}{du}=dfrac{dv}{du}+1+dfrac{1}{(u^2+1)^frac{3}{2}}$



          $therefore vleft(dfrac{dv}{du}+1+dfrac{1}{(u^2+1)^frac{3}{2}}right)=v+u+dfrac{u}{sqrt{u^2+1}}$



          $vdfrac{dv}{du}+left(1+dfrac{1}{(u^2+1)^frac{3}{2}}right)v=v+u+dfrac{u}{sqrt{u^2+1}}$



          $vdfrac{dv}{du}=-dfrac{v}{(u^2+1)^frac{3}{2}}+u+dfrac{u}{sqrt{u^2+1}}$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Very nice but still i have no clue to proceed from the last step.
            $endgroup$
            – Umesh shankar
            Jan 31 at 14:45














          2












          2








          2





          $begingroup$

          Hint:



          Let $x=yu$ ,



          Then $dfrac{dx}{dy}=ydfrac{du}{dy}+u$



          $therefore ydfrac{du}{dy}+u+dfrac{yu}{sqrt{y^2u^2+y^2}}=y$



          $ydfrac{du}{dy}+u+dfrac{u}{sqrt{u^2+1}}=y$



          $ydfrac{du}{dy}=y-u-dfrac{u}{sqrt{u^2+1}}$



          $left(y-u-dfrac{u}{sqrt{u^2+1}}right)dfrac{dy}{du}=y$



          This belongs to an Abel equation of the second kind.



          Let $v=y-u-dfrac{u}{sqrt{u^2+1}}$ ,



          Then $y=v+u+dfrac{u}{sqrt{u^2+1}}$



          $dfrac{dy}{du}=dfrac{dv}{du}+1+dfrac{1}{(u^2+1)^frac{3}{2}}$



          $therefore vleft(dfrac{dv}{du}+1+dfrac{1}{(u^2+1)^frac{3}{2}}right)=v+u+dfrac{u}{sqrt{u^2+1}}$



          $vdfrac{dv}{du}+left(1+dfrac{1}{(u^2+1)^frac{3}{2}}right)v=v+u+dfrac{u}{sqrt{u^2+1}}$



          $vdfrac{dv}{du}=-dfrac{v}{(u^2+1)^frac{3}{2}}+u+dfrac{u}{sqrt{u^2+1}}$






          share|cite|improve this answer









          $endgroup$



          Hint:



          Let $x=yu$ ,



          Then $dfrac{dx}{dy}=ydfrac{du}{dy}+u$



          $therefore ydfrac{du}{dy}+u+dfrac{yu}{sqrt{y^2u^2+y^2}}=y$



          $ydfrac{du}{dy}+u+dfrac{u}{sqrt{u^2+1}}=y$



          $ydfrac{du}{dy}=y-u-dfrac{u}{sqrt{u^2+1}}$



          $left(y-u-dfrac{u}{sqrt{u^2+1}}right)dfrac{dy}{du}=y$



          This belongs to an Abel equation of the second kind.



          Let $v=y-u-dfrac{u}{sqrt{u^2+1}}$ ,



          Then $y=v+u+dfrac{u}{sqrt{u^2+1}}$



          $dfrac{dy}{du}=dfrac{dv}{du}+1+dfrac{1}{(u^2+1)^frac{3}{2}}$



          $therefore vleft(dfrac{dv}{du}+1+dfrac{1}{(u^2+1)^frac{3}{2}}right)=v+u+dfrac{u}{sqrt{u^2+1}}$



          $vdfrac{dv}{du}+left(1+dfrac{1}{(u^2+1)^frac{3}{2}}right)v=v+u+dfrac{u}{sqrt{u^2+1}}$



          $vdfrac{dv}{du}=-dfrac{v}{(u^2+1)^frac{3}{2}}+u+dfrac{u}{sqrt{u^2+1}}$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 31 at 12:12









          doraemonpauldoraemonpaul

          12.8k31661




          12.8k31661












          • $begingroup$
            Very nice but still i have no clue to proceed from the last step.
            $endgroup$
            – Umesh shankar
            Jan 31 at 14:45


















          • $begingroup$
            Very nice but still i have no clue to proceed from the last step.
            $endgroup$
            – Umesh shankar
            Jan 31 at 14:45
















          $begingroup$
          Very nice but still i have no clue to proceed from the last step.
          $endgroup$
          – Umesh shankar
          Jan 31 at 14:45




          $begingroup$
          Very nice but still i have no clue to proceed from the last step.
          $endgroup$
          – Umesh shankar
          Jan 31 at 14:45


















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