Resolution of differential equation











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All functions verify the condition below :



$$f''(x)+f(-x)=x+cos(x)$$



I have to use something related to differential equations.






real-analysis differential-equations







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    up vote
    2
    down vote

    favorite












    All functions verify the condition below :



    $$f''(x)+f(-x)=x+cos(x)$$



    I have to use something related to differential equations.






    real-analysis differential-equations







    share|cite|improve this question


























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      All functions verify the condition below :



      $$f''(x)+f(-x)=x+cos(x)$$



      I have to use something related to differential equations.






      real-analysis differential-equations







      share|cite|improve this question















      All functions verify the condition below :



      $$f''(x)+f(-x)=x+cos(x)$$



      I have to use something related to differential equations.






      real-analysis differential-equations




      real-analysis differential-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago









      user376343

      2,7012822




      2,7012822










      asked 2 hours ago









      ahmed bennani

      193




      193






















          2 Answers
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          down vote













          Any real function can be written in a unique way as the sum of an even function and an odd function, say
          $$f=g+h$$
          where $g$ is even and $h$ is odd. It is easy to check that $g''$ is also even and $h''$ is also odd. So we have
          $$g''(x)+h''(x)+g(x)-h(x)=x+cos x .$$
          Equating even and odd parts,
          $$g''(x)+g(x)=cos x ,qquad h''(x)-h(x)=x .$$
          Solving by standard methods,
          $$g(x)=Acos x+Bsin x+tfrac12xsin x ,qquad h(x)=Ccosh x+Dsinh x-x .$$
          But $g$ is even so $B=0$, and $h$ is odd so $C=0$. Hence
          $$f(x)=g(x)+h(x)=Acos x+Dsinh x+tfrac12xsin x-x .$$






          share|cite|improve this answer




























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            Iterating the equation you get $f^{(4)}(x)+f(x)=-x$ which can be solved by standard methods. Solutions of $lambda^{4}+1=0$ are $frac {pm 1 pm i} {sqrt 2}$. and when you go back to the original DE you find that only two of these values give solutions of the original DE. I hope you can fill in the details.






            share|cite|improve this answer























            • This introduces extra solutions which are not solutions of the given equation.
              – David
              2 hours ago










            • You get $f^{IV}(x)+f''(-x)=-cos(x)$ and $f''(-x)+f(x)=-x+cos(x)$, therefore $$f^{IV}(x)-f(x)=x-2cos(x),$$ which is different from what you state.
              – LutzL
              13 mins ago













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            2 Answers
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            2 Answers
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            up vote
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            Any real function can be written in a unique way as the sum of an even function and an odd function, say
            $$f=g+h$$
            where $g$ is even and $h$ is odd. It is easy to check that $g''$ is also even and $h''$ is also odd. So we have
            $$g''(x)+h''(x)+g(x)-h(x)=x+cos x .$$
            Equating even and odd parts,
            $$g''(x)+g(x)=cos x ,qquad h''(x)-h(x)=x .$$
            Solving by standard methods,
            $$g(x)=Acos x+Bsin x+tfrac12xsin x ,qquad h(x)=Ccosh x+Dsinh x-x .$$
            But $g$ is even so $B=0$, and $h$ is odd so $C=0$. Hence
            $$f(x)=g(x)+h(x)=Acos x+Dsinh x+tfrac12xsin x-x .$$






            share|cite|improve this answer

























              up vote
              5
              down vote













              Any real function can be written in a unique way as the sum of an even function and an odd function, say
              $$f=g+h$$
              where $g$ is even and $h$ is odd. It is easy to check that $g''$ is also even and $h''$ is also odd. So we have
              $$g''(x)+h''(x)+g(x)-h(x)=x+cos x .$$
              Equating even and odd parts,
              $$g''(x)+g(x)=cos x ,qquad h''(x)-h(x)=x .$$
              Solving by standard methods,
              $$g(x)=Acos x+Bsin x+tfrac12xsin x ,qquad h(x)=Ccosh x+Dsinh x-x .$$
              But $g$ is even so $B=0$, and $h$ is odd so $C=0$. Hence
              $$f(x)=g(x)+h(x)=Acos x+Dsinh x+tfrac12xsin x-x .$$






              share|cite|improve this answer























                up vote
                5
                down vote










                up vote
                5
                down vote









                Any real function can be written in a unique way as the sum of an even function and an odd function, say
                $$f=g+h$$
                where $g$ is even and $h$ is odd. It is easy to check that $g''$ is also even and $h''$ is also odd. So we have
                $$g''(x)+h''(x)+g(x)-h(x)=x+cos x .$$
                Equating even and odd parts,
                $$g''(x)+g(x)=cos x ,qquad h''(x)-h(x)=x .$$
                Solving by standard methods,
                $$g(x)=Acos x+Bsin x+tfrac12xsin x ,qquad h(x)=Ccosh x+Dsinh x-x .$$
                But $g$ is even so $B=0$, and $h$ is odd so $C=0$. Hence
                $$f(x)=g(x)+h(x)=Acos x+Dsinh x+tfrac12xsin x-x .$$






                share|cite|improve this answer












                Any real function can be written in a unique way as the sum of an even function and an odd function, say
                $$f=g+h$$
                where $g$ is even and $h$ is odd. It is easy to check that $g''$ is also even and $h''$ is also odd. So we have
                $$g''(x)+h''(x)+g(x)-h(x)=x+cos x .$$
                Equating even and odd parts,
                $$g''(x)+g(x)=cos x ,qquad h''(x)-h(x)=x .$$
                Solving by standard methods,
                $$g(x)=Acos x+Bsin x+tfrac12xsin x ,qquad h(x)=Ccosh x+Dsinh x-x .$$
                But $g$ is even so $B=0$, and $h$ is odd so $C=0$. Hence
                $$f(x)=g(x)+h(x)=Acos x+Dsinh x+tfrac12xsin x-x .$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                David

                67.4k663126




                67.4k663126






















                    up vote
                    0
                    down vote













                    Iterating the equation you get $f^{(4)}(x)+f(x)=-x$ which can be solved by standard methods. Solutions of $lambda^{4}+1=0$ are $frac {pm 1 pm i} {sqrt 2}$. and when you go back to the original DE you find that only two of these values give solutions of the original DE. I hope you can fill in the details.






                    share|cite|improve this answer























                    • This introduces extra solutions which are not solutions of the given equation.
                      – David
                      2 hours ago










                    • You get $f^{IV}(x)+f''(-x)=-cos(x)$ and $f''(-x)+f(x)=-x+cos(x)$, therefore $$f^{IV}(x)-f(x)=x-2cos(x),$$ which is different from what you state.
                      – LutzL
                      13 mins ago

















                    up vote
                    0
                    down vote













                    Iterating the equation you get $f^{(4)}(x)+f(x)=-x$ which can be solved by standard methods. Solutions of $lambda^{4}+1=0$ are $frac {pm 1 pm i} {sqrt 2}$. and when you go back to the original DE you find that only two of these values give solutions of the original DE. I hope you can fill in the details.






                    share|cite|improve this answer























                    • This introduces extra solutions which are not solutions of the given equation.
                      – David
                      2 hours ago










                    • You get $f^{IV}(x)+f''(-x)=-cos(x)$ and $f''(-x)+f(x)=-x+cos(x)$, therefore $$f^{IV}(x)-f(x)=x-2cos(x),$$ which is different from what you state.
                      – LutzL
                      13 mins ago















                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    Iterating the equation you get $f^{(4)}(x)+f(x)=-x$ which can be solved by standard methods. Solutions of $lambda^{4}+1=0$ are $frac {pm 1 pm i} {sqrt 2}$. and when you go back to the original DE you find that only two of these values give solutions of the original DE. I hope you can fill in the details.






                    share|cite|improve this answer














                    Iterating the equation you get $f^{(4)}(x)+f(x)=-x$ which can be solved by standard methods. Solutions of $lambda^{4}+1=0$ are $frac {pm 1 pm i} {sqrt 2}$. and when you go back to the original DE you find that only two of these values give solutions of the original DE. I hope you can fill in the details.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 2 hours ago

























                    answered 2 hours ago









                    Kavi Rama Murthy

                    46.6k31854




                    46.6k31854












                    • This introduces extra solutions which are not solutions of the given equation.
                      – David
                      2 hours ago










                    • You get $f^{IV}(x)+f''(-x)=-cos(x)$ and $f''(-x)+f(x)=-x+cos(x)$, therefore $$f^{IV}(x)-f(x)=x-2cos(x),$$ which is different from what you state.
                      – LutzL
                      13 mins ago




















                    • This introduces extra solutions which are not solutions of the given equation.
                      – David
                      2 hours ago










                    • You get $f^{IV}(x)+f''(-x)=-cos(x)$ and $f''(-x)+f(x)=-x+cos(x)$, therefore $$f^{IV}(x)-f(x)=x-2cos(x),$$ which is different from what you state.
                      – LutzL
                      13 mins ago


















                    This introduces extra solutions which are not solutions of the given equation.
                    – David
                    2 hours ago




                    This introduces extra solutions which are not solutions of the given equation.
                    – David
                    2 hours ago












                    You get $f^{IV}(x)+f''(-x)=-cos(x)$ and $f''(-x)+f(x)=-x+cos(x)$, therefore $$f^{IV}(x)-f(x)=x-2cos(x),$$ which is different from what you state.
                    – LutzL
                    13 mins ago






                    You get $f^{IV}(x)+f''(-x)=-cos(x)$ and $f''(-x)+f(x)=-x+cos(x)$, therefore $$f^{IV}(x)-f(x)=x-2cos(x),$$ which is different from what you state.
                    – LutzL
                    13 mins ago




















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