Homogeneous or non - homogeneous $?$












2












$begingroup$


The second order differential equation is given by -



$ frac{d^{2}y}{dx^{2}} + sin (x+y) = sin x$



Is this a homogeneous differential equation $?$



Well, I guess this is not a homogeneous differential equation since the form of this equation is not $a(x)y'' + b(x)y' +c(x)y = 0$.
But the answer is given that it's homogeneous.
How can this equation be homogeneous?






ordinary-differential-equations







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    The second order differential equation is given by -



    $ frac{d^{2}y}{dx^{2}} + sin (x+y) = sin x$



    Is this a homogeneous differential equation $?$



    Well, I guess this is not a homogeneous differential equation since the form of this equation is not $a(x)y'' + b(x)y' +c(x)y = 0$.
    But the answer is given that it's homogeneous.
    How can this equation be homogeneous?






    ordinary-differential-equations







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      2



      $begingroup$


      The second order differential equation is given by -



      $ frac{d^{2}y}{dx^{2}} + sin (x+y) = sin x$



      Is this a homogeneous differential equation $?$



      Well, I guess this is not a homogeneous differential equation since the form of this equation is not $a(x)y'' + b(x)y' +c(x)y = 0$.
      But the answer is given that it's homogeneous.
      How can this equation be homogeneous?






      ordinary-differential-equations







      share|cite|improve this question









      $endgroup$




      The second order differential equation is given by -



      $ frac{d^{2}y}{dx^{2}} + sin (x+y) = sin x$



      Is this a homogeneous differential equation $?$



      Well, I guess this is not a homogeneous differential equation since the form of this equation is not $a(x)y'' + b(x)y' +c(x)y = 0$.
      But the answer is given that it's homogeneous.
      How can this equation be homogeneous?






      ordinary-differential-equations




      ordinary-differential-equations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 24 at 18:33









      MathsaddictMathsaddict

      3669




      3669






















          2 Answers
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          0












          $begingroup$

          Let $u=x+y$ ,



          Then $dfrac{du}{dx}=1+dfrac{dy}{dx}$



          $dfrac{d^2u}{dx^2}=dfrac{d^2y}{dx^2}$



          $thereforedfrac{d^2u}{dx^2}+sin u=sin x$



          This is a inhomogeneous differential equation.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            You are correct, as it is not a linear ODE, it is neither homogeneous nor inhomogeneous.



            The cited characterization is most likely based on the fact that $y=0$ is a solution, but that is only a necessary condition for linearity, not a sufficient one.






            share|cite|improve this answer









            $endgroup$













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              2 Answers
              2






              active

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              2 Answers
              2






              active

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              active

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              active

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              0












              $begingroup$

              Let $u=x+y$ ,



              Then $dfrac{du}{dx}=1+dfrac{dy}{dx}$



              $dfrac{d^2u}{dx^2}=dfrac{d^2y}{dx^2}$



              $thereforedfrac{d^2u}{dx^2}+sin u=sin x$



              This is a inhomogeneous differential equation.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Let $u=x+y$ ,



                Then $dfrac{du}{dx}=1+dfrac{dy}{dx}$



                $dfrac{d^2u}{dx^2}=dfrac{d^2y}{dx^2}$



                $thereforedfrac{d^2u}{dx^2}+sin u=sin x$



                This is a inhomogeneous differential equation.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Let $u=x+y$ ,



                  Then $dfrac{du}{dx}=1+dfrac{dy}{dx}$



                  $dfrac{d^2u}{dx^2}=dfrac{d^2y}{dx^2}$



                  $thereforedfrac{d^2u}{dx^2}+sin u=sin x$



                  This is a inhomogeneous differential equation.






                  share|cite|improve this answer









                  $endgroup$



                  Let $u=x+y$ ,



                  Then $dfrac{du}{dx}=1+dfrac{dy}{dx}$



                  $dfrac{d^2u}{dx^2}=dfrac{d^2y}{dx^2}$



                  $thereforedfrac{d^2u}{dx^2}+sin u=sin x$



                  This is a inhomogeneous differential equation.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 25 at 12:39









                  doraemonpauldoraemonpaul

                  12.7k31660




                  12.7k31660























                      0












                      $begingroup$

                      You are correct, as it is not a linear ODE, it is neither homogeneous nor inhomogeneous.



                      The cited characterization is most likely based on the fact that $y=0$ is a solution, but that is only a necessary condition for linearity, not a sufficient one.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        You are correct, as it is not a linear ODE, it is neither homogeneous nor inhomogeneous.



                        The cited characterization is most likely based on the fact that $y=0$ is a solution, but that is only a necessary condition for linearity, not a sufficient one.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          You are correct, as it is not a linear ODE, it is neither homogeneous nor inhomogeneous.



                          The cited characterization is most likely based on the fact that $y=0$ is a solution, but that is only a necessary condition for linearity, not a sufficient one.






                          share|cite|improve this answer









                          $endgroup$



                          You are correct, as it is not a linear ODE, it is neither homogeneous nor inhomogeneous.



                          The cited characterization is most likely based on the fact that $y=0$ is a solution, but that is only a necessary condition for linearity, not a sufficient one.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 25 at 13:00









                          LutzLLutzL

                          59.6k42057




                          59.6k42057






























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