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

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How would I solve an equation where one of the differential terms is to a power? For example:
$frac{d^2y}{dx^2}+k(frac{dy}{dx})^2=0$?



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






calculus ordinary-differential-equations







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    $begingroup$


    How would I solve an equation where one of the differential terms is to a power? For example:
    $frac{d^2y}{dx^2}+k(frac{dy}{dx})^2=0$?



    I've been given advice to use the $D$ operator which apparently means $frac{d}{dx}()$ 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.







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      4












      4








      4





      $begingroup$


      How would I solve an equation where one of the differential terms is to a power? For example:
      $frac{d^2y}{dx^2}+k(frac{dy}{dx})^2=0$?



      I've been given advice to use the $D$ operator which apparently means $frac{d}{dx}()$ 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.







      $endgroup$




      How would I solve an equation where one of the differential terms is to a power? For example:
      $frac{d^2y}{dx^2}+k(frac{dy}{dx})^2=0$?



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






      calculus ordinary-differential-equations




      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.









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









      asked 10 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 $v=frac{dy}{dx}$. Your differential equation will turn into $frac{dv}{dx}+kv^2=0$, and I guess you will manage to take it from here.






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            1 Answer
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            active

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

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            active

            oldest

            votes









            10












            $begingroup$

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






            share|cite|improve this answer









            $endgroup$


















              10












              $begingroup$

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






              share|cite|improve this answer









              $endgroup$
















                10












                10








                10





                $begingroup$

                Since you only have second and first derivatives of $y$ and no
                un-differentiated $y$, you could try to introduce the new function $v=frac{dy}{dx}$. Your differential equation will turn into $frac{dv}{dx}+kv^2=0$, 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 $v=frac{dy}{dx}$. Your differential equation will turn into $frac{dv}{dx}+kv^2=0$, 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 10 hours ago









                mickepmickep

                18.7k12251




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