Euler-Ansatz for hom. ODEs with constant coefficients











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Given is an hom. ODE with constant constant coefficients:



$A_0y(x)+A_1y'(x)+A_2y''(x) + dots + A_ny^{(n)}=0 tag{1}$



Now it's clear to me that the solution space $yinmathbb L$ is a vector space.



We can solve (1) using the Ansatz: $y(x)=e^{kx}$. We get:



$chi(k)=A_0 + A_1k + A_2k^2 + dots + A_nk^n=0 tag{2}$



Now with $k_i$ being a solution to $chi$ with multiplicity $m_i$, the solution to the ODE is:



$y(x)=sum_i y_i(x) tag{3}$



with



$y_i(x)=sum_{j=0}^{m_i} C_j x^j e^{k_i x}, quad C_jinmathbb R tag{4}$



Question: Since I expect $mathbb L$ to be a vector space, (4) kind of makes sense. I mean it doesn't look wrong and I can work with it and solve such ODEs, but I can't derive it. So (1), (2), (3) is clear but (4) isn't. How exactly do we get the $x^j$ part in (4)?






calculus differential-equations







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    Given is an hom. ODE with constant constant coefficients:



    $A_0y(x)+A_1y'(x)+A_2y''(x) + dots + A_ny^{(n)}=0 tag{1}$



    Now it's clear to me that the solution space $yinmathbb L$ is a vector space.



    We can solve (1) using the Ansatz: $y(x)=e^{kx}$. We get:



    $chi(k)=A_0 + A_1k + A_2k^2 + dots + A_nk^n=0 tag{2}$



    Now with $k_i$ being a solution to $chi$ with multiplicity $m_i$, the solution to the ODE is:



    $y(x)=sum_i y_i(x) tag{3}$



    with



    $y_i(x)=sum_{j=0}^{m_i} C_j x^j e^{k_i x}, quad C_jinmathbb R tag{4}$



    Question: Since I expect $mathbb L$ to be a vector space, (4) kind of makes sense. I mean it doesn't look wrong and I can work with it and solve such ODEs, but I can't derive it. So (1), (2), (3) is clear but (4) isn't. How exactly do we get the $x^j$ part in (4)?






    calculus differential-equations







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Given is an hom. ODE with constant constant coefficients:



      $A_0y(x)+A_1y'(x)+A_2y''(x) + dots + A_ny^{(n)}=0 tag{1}$



      Now it's clear to me that the solution space $yinmathbb L$ is a vector space.



      We can solve (1) using the Ansatz: $y(x)=e^{kx}$. We get:



      $chi(k)=A_0 + A_1k + A_2k^2 + dots + A_nk^n=0 tag{2}$



      Now with $k_i$ being a solution to $chi$ with multiplicity $m_i$, the solution to the ODE is:



      $y(x)=sum_i y_i(x) tag{3}$



      with



      $y_i(x)=sum_{j=0}^{m_i} C_j x^j e^{k_i x}, quad C_jinmathbb R tag{4}$



      Question: Since I expect $mathbb L$ to be a vector space, (4) kind of makes sense. I mean it doesn't look wrong and I can work with it and solve such ODEs, but I can't derive it. So (1), (2), (3) is clear but (4) isn't. How exactly do we get the $x^j$ part in (4)?






      calculus differential-equations







      share|cite|improve this question













      Given is an hom. ODE with constant constant coefficients:



      $A_0y(x)+A_1y'(x)+A_2y''(x) + dots + A_ny^{(n)}=0 tag{1}$



      Now it's clear to me that the solution space $yinmathbb L$ is a vector space.



      We can solve (1) using the Ansatz: $y(x)=e^{kx}$. We get:



      $chi(k)=A_0 + A_1k + A_2k^2 + dots + A_nk^n=0 tag{2}$



      Now with $k_i$ being a solution to $chi$ with multiplicity $m_i$, the solution to the ODE is:



      $y(x)=sum_i y_i(x) tag{3}$



      with



      $y_i(x)=sum_{j=0}^{m_i} C_j x^j e^{k_i x}, quad C_jinmathbb R tag{4}$



      Question: Since I expect $mathbb L$ to be a vector space, (4) kind of makes sense. I mean it doesn't look wrong and I can work with it and solve such ODEs, but I can't derive it. So (1), (2), (3) is clear but (4) isn't. How exactly do we get the $x^j$ part in (4)?






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          Defining $v = (y, y^{(1)}, ..., y^{(n-1)})$, you can interpret your ODE as a linear system of differential equations,



          $frac{dv}{dt} = Av$



          $v(0) = v_0$



          where $A$ is an appropriate constant matrix. Now, you can show that



          $v(t) = (I + At + frac{(At)^2}{2!} + ...)v_0$



          is a well defined analytic solution of this problem (in fact, the only solution). This infinite sum defines the exponential of a matrix, $e^{At}$.



          Example: $y'' - 2y' + y = 0$. Defining $v = (y, y')$, we have



          $v' = begin{pmatrix} 0 & 1 \ -1 & 2 end{pmatrix} v$



          The characteristic polynomial of this matrix (and of the associated ODE) has only one root, and for that reason the exponential of $At$ will be something of the form



          $e^{At} = Qe^tbegin{pmatrix} 1 & 0 \ t & 1 end{pmatrix}Q^{-1}$, where Q is some constant matrix (search for Jordan Decomposition, in the context of the exponential of a matrix). And that's where the $t^j$ will come from :)






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            up vote
            1
            down vote



            accepted










            Defining $v = (y, y^{(1)}, ..., y^{(n-1)})$, you can interpret your ODE as a linear system of differential equations,



            $frac{dv}{dt} = Av$



            $v(0) = v_0$



            where $A$ is an appropriate constant matrix. Now, you can show that



            $v(t) = (I + At + frac{(At)^2}{2!} + ...)v_0$



            is a well defined analytic solution of this problem (in fact, the only solution). This infinite sum defines the exponential of a matrix, $e^{At}$.



            Example: $y'' - 2y' + y = 0$. Defining $v = (y, y')$, we have



            $v' = begin{pmatrix} 0 & 1 \ -1 & 2 end{pmatrix} v$



            The characteristic polynomial of this matrix (and of the associated ODE) has only one root, and for that reason the exponential of $At$ will be something of the form



            $e^{At} = Qe^tbegin{pmatrix} 1 & 0 \ t & 1 end{pmatrix}Q^{-1}$, where Q is some constant matrix (search for Jordan Decomposition, in the context of the exponential of a matrix). And that's where the $t^j$ will come from :)






            share|cite|improve this answer



























              up vote
              1
              down vote



              accepted










              Defining $v = (y, y^{(1)}, ..., y^{(n-1)})$, you can interpret your ODE as a linear system of differential equations,



              $frac{dv}{dt} = Av$



              $v(0) = v_0$



              where $A$ is an appropriate constant matrix. Now, you can show that



              $v(t) = (I + At + frac{(At)^2}{2!} + ...)v_0$



              is a well defined analytic solution of this problem (in fact, the only solution). This infinite sum defines the exponential of a matrix, $e^{At}$.



              Example: $y'' - 2y' + y = 0$. Defining $v = (y, y')$, we have



              $v' = begin{pmatrix} 0 & 1 \ -1 & 2 end{pmatrix} v$



              The characteristic polynomial of this matrix (and of the associated ODE) has only one root, and for that reason the exponential of $At$ will be something of the form



              $e^{At} = Qe^tbegin{pmatrix} 1 & 0 \ t & 1 end{pmatrix}Q^{-1}$, where Q is some constant matrix (search for Jordan Decomposition, in the context of the exponential of a matrix). And that's where the $t^j$ will come from :)






              share|cite|improve this answer

























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Defining $v = (y, y^{(1)}, ..., y^{(n-1)})$, you can interpret your ODE as a linear system of differential equations,



                $frac{dv}{dt} = Av$



                $v(0) = v_0$



                where $A$ is an appropriate constant matrix. Now, you can show that



                $v(t) = (I + At + frac{(At)^2}{2!} + ...)v_0$



                is a well defined analytic solution of this problem (in fact, the only solution). This infinite sum defines the exponential of a matrix, $e^{At}$.



                Example: $y'' - 2y' + y = 0$. Defining $v = (y, y')$, we have



                $v' = begin{pmatrix} 0 & 1 \ -1 & 2 end{pmatrix} v$



                The characteristic polynomial of this matrix (and of the associated ODE) has only one root, and for that reason the exponential of $At$ will be something of the form



                $e^{At} = Qe^tbegin{pmatrix} 1 & 0 \ t & 1 end{pmatrix}Q^{-1}$, where Q is some constant matrix (search for Jordan Decomposition, in the context of the exponential of a matrix). And that's where the $t^j$ will come from :)






                share|cite|improve this answer














                Defining $v = (y, y^{(1)}, ..., y^{(n-1)})$, you can interpret your ODE as a linear system of differential equations,



                $frac{dv}{dt} = Av$



                $v(0) = v_0$



                where $A$ is an appropriate constant matrix. Now, you can show that



                $v(t) = (I + At + frac{(At)^2}{2!} + ...)v_0$



                is a well defined analytic solution of this problem (in fact, the only solution). This infinite sum defines the exponential of a matrix, $e^{At}$.



                Example: $y'' - 2y' + y = 0$. Defining $v = (y, y')$, we have



                $v' = begin{pmatrix} 0 & 1 \ -1 & 2 end{pmatrix} v$



                The characteristic polynomial of this matrix (and of the associated ODE) has only one root, and for that reason the exponential of $At$ will be something of the form



                $e^{At} = Qe^tbegin{pmatrix} 1 & 0 \ t & 1 end{pmatrix}Q^{-1}$, where Q is some constant matrix (search for Jordan Decomposition, in the context of the exponential of a matrix). And that's where the $t^j$ will come from :)







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








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