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How to solve a differential equation with a term to a power?



The Next CEO of Stack OverflowHow do I solve this differential equation?Solving differential equation $x^2y''-xy'+y=0, x>0$ with non-constant coefficients using characteristic equation?Solution to differential equationCan I solve an Euler differential equation by using the Frobenius method?How to solve a differential equationHelp needed with differential equationHow to solve differential equations that look like theseHow to solve differential equation with one differential termsolve differential equation $fracdPdt = kPcos^2(rt-Theta)$How to solve an exponential differential equation?










2












$begingroup$


How would I solve an equation where one of the differential terms is to a power? For example:
$fracd^2ydx^2+k(fracdydx)^2=0$?



I've been given advice to use the $D$ operator which apparently means $fracddx()$ but I'm not sure how that's applicable to this scenario. Any alternative suggestions or explanations would be appreciated!










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Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
















    2












    $begingroup$


    How would I solve an equation where one of the differential terms is to a power? For example:
    $fracd^2ydx^2+k(fracdydx)^2=0$?



    I've been given advice to use the $D$ operator which apparently means $fracddx()$ but I'm not sure how that's applicable to this scenario. Any alternative suggestions or explanations would be appreciated!










    share|cite|improve this question







    New contributor




    Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$














      2












      2








      2





      $begingroup$


      How would I solve an equation where one of the differential terms is to a power? For example:
      $fracd^2ydx^2+k(fracdydx)^2=0$?



      I've been given advice to use the $D$ operator which apparently means $fracddx()$ but I'm not sure how that's applicable to this scenario. Any alternative suggestions or explanations would be appreciated!










      share|cite|improve this question







      New contributor




      Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      How would I solve an equation where one of the differential terms is to a power? For example:
      $fracd^2ydx^2+k(fracdydx)^2=0$?



      I've been given advice to use the $D$ operator which apparently means $fracddx()$ but I'm not sure how that's applicable to this scenario. Any alternative suggestions or explanations would be appreciated!







      calculus ordinary-differential-equations






      share|cite|improve this question







      New contributor




      Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question






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      Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 4 hours ago









      Ammar TarajiaAmmar Tarajia

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      Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          $begingroup$

          Since you only have second and first derivatives of $y$ and no
          un-differentiated $y$, you could try to introduce the new function $v=fracdydx$. Your differential equation will turn into $fracdvdx+kv^2=0$, and I guess you will manage to take it from here.






          share|cite|improve this answer









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            5












            $begingroup$

            Since you only have second and first derivatives of $y$ and no
            un-differentiated $y$, you could try to introduce the new function $v=fracdydx$. Your differential equation will turn into $fracdvdx+kv^2=0$, and I guess you will manage to take it from here.






            share|cite|improve this answer









            $endgroup$

















              5












              $begingroup$

              Since you only have second and first derivatives of $y$ and no
              un-differentiated $y$, you could try to introduce the new function $v=fracdydx$. Your differential equation will turn into $fracdvdx+kv^2=0$, and I guess you will manage to take it from here.






              share|cite|improve this answer









              $endgroup$















                5












                5








                5





                $begingroup$

                Since you only have second and first derivatives of $y$ and no
                un-differentiated $y$, you could try to introduce the new function $v=fracdydx$. Your differential equation will turn into $fracdvdx+kv^2=0$, and I guess you will manage to take it from here.






                share|cite|improve this answer









                $endgroup$



                Since you only have second and first derivatives of $y$ and no
                un-differentiated $y$, you could try to introduce the new function $v=fracdydx$. Your differential equation will turn into $fracdvdx+kv^2=0$, and I guess you will manage to take it from here.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 4 hours ago









                mickepmickep

                18.7k12250




                18.7k12250




















                    Ammar Tarajia is a new contributor. Be nice, and check out our Code of Conduct.









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