How to find solution to a non-homogenous system of differential equations












0












$begingroup$



Let $X'=AX + b(t)$ be a system of differential equations where: $X(t) = (x_1(t),...,x_n(t))$, $Ainmathcal M_n({mathbb{R}})$ and $b(t) = (b_1(t),..,b_n(t))$.




How do I solve this system using the variations of constants method?



I know that if $M(t)=(x_1(t)|....|x_n(t))$ a matrix of fundamental solutions and



$$X(t)=c_1(t)x_1(t)+...+c_n(t)x_n(t)=M(t)C(t)$$



where $C(t) = (c_1(t)|..|c_n(t))$



And that $C(t)$ verifies $C'(t)=M^{-1}(t)b(t)$



If I have an example like:



$$left{begin{array}{l}x'=x+y-cos t\y'=-2x-y+sin t+cos tend{array}right.$$



How do I actually solve it?






ordinary-differential-equations systems-of-equations







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    0












    $begingroup$



    Let $X'=AX + b(t)$ be a system of differential equations where: $X(t) = (x_1(t),...,x_n(t))$, $Ainmathcal M_n({mathbb{R}})$ and $b(t) = (b_1(t),..,b_n(t))$.




    How do I solve this system using the variations of constants method?



    I know that if $M(t)=(x_1(t)|....|x_n(t))$ a matrix of fundamental solutions and



    $$X(t)=c_1(t)x_1(t)+...+c_n(t)x_n(t)=M(t)C(t)$$



    where $C(t) = (c_1(t)|..|c_n(t))$



    And that $C(t)$ verifies $C'(t)=M^{-1}(t)b(t)$



    If I have an example like:



    $$left{begin{array}{l}x'=x+y-cos t\y'=-2x-y+sin t+cos tend{array}right.$$



    How do I actually solve it?






    ordinary-differential-equations systems-of-equations







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$



      Let $X'=AX + b(t)$ be a system of differential equations where: $X(t) = (x_1(t),...,x_n(t))$, $Ainmathcal M_n({mathbb{R}})$ and $b(t) = (b_1(t),..,b_n(t))$.




      How do I solve this system using the variations of constants method?



      I know that if $M(t)=(x_1(t)|....|x_n(t))$ a matrix of fundamental solutions and



      $$X(t)=c_1(t)x_1(t)+...+c_n(t)x_n(t)=M(t)C(t)$$



      where $C(t) = (c_1(t)|..|c_n(t))$



      And that $C(t)$ verifies $C'(t)=M^{-1}(t)b(t)$



      If I have an example like:



      $$left{begin{array}{l}x'=x+y-cos t\y'=-2x-y+sin t+cos tend{array}right.$$



      How do I actually solve it?






      ordinary-differential-equations systems-of-equations







      share|cite|improve this question









      $endgroup$





      Let $X'=AX + b(t)$ be a system of differential equations where: $X(t) = (x_1(t),...,x_n(t))$, $Ainmathcal M_n({mathbb{R}})$ and $b(t) = (b_1(t),..,b_n(t))$.




      How do I solve this system using the variations of constants method?



      I know that if $M(t)=(x_1(t)|....|x_n(t))$ a matrix of fundamental solutions and



      $$X(t)=c_1(t)x_1(t)+...+c_n(t)x_n(t)=M(t)C(t)$$



      where $C(t) = (c_1(t)|..|c_n(t))$



      And that $C(t)$ verifies $C'(t)=M^{-1}(t)b(t)$



      If I have an example like:



      $$left{begin{array}{l}x'=x+y-cos t\y'=-2x-y+sin t+cos tend{array}right.$$



      How do I actually solve it?






      ordinary-differential-equations systems-of-equations




      ordinary-differential-equations systems-of-equations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 14 at 14:43









      C. CristiC. Cristi

      1,629218




      1,629218






















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          $begingroup$

          You were almost done, as you can simply integrate the function $M^{-1}(t)b(t)$ over $t$ to obtain the function $C(t)$, up to constants of integration.



          You write your example in the form
          begin{equation}
          boldsymbol{dot{x}} = left( begin{array}{c}
          dot{x}\
          dot{y}
          end{array}
          right) = left( begin{array}{cc}
          1 & 1\
          -2 & -1
          end{array}
          right) left( begin{array}{c}
          x\
          y
          end{array}
          right) + left( begin{array}{c}
          -cos(t)\
          sin(t)+cos(t)
          end{array}
          right) = boldsymbol{underline{A}} boldsymbol{x} + boldsymbol{b}(t),
          end{equation}
          which has the general solution
          begin{equation}
          boldsymbol{x}(t) = boldsymbol{underline{M}}(t) int boldsymbol{underline{M}}^{-1}(t) boldsymbol{b}(t) , mathrm{d}t,
          end{equation}
          where $boldsymbol{underline{M}}(t) := e^{t boldsymbol{underline{A}}}$ is a matrix exponential (with inverse $boldsymbol{underline{M}}^{-1}(t) = e^{-t boldsymbol{underline{A}}}$). The constants of integration are contained in the indefinite integral. We obtain
          begin{equation}
          boldsymbol{underline{M}}(t) = e^{tboldsymbol{underline{A}}} = left( begin{array}{cc}
          cos(t) + sin(t) & sin(t)\
          -2sin(t) & cos(t) - sin(t)
          end{array}
          right), quad int boldsymbol{underline{M}}^{-1}(t) boldsymbol{b}(t) , mathrm{d}t = left( begin{array}{c}
          C_1 - t\
          C_2 + t
          end{array}
          right),
          end{equation}
          $C_1, C_2 in mathbb{R}$, and finally
          begin{equation}
          boldsymbol{x}(t) = left( begin{array}{c}
          x(t)\
          y(t)
          end{array}
          right) = left( begin{array}{c}
          -t cos(t) + C_1 (cos(t) + sin(t)) + C_2 sin(t)\
          t cos(t) + t sin(t) - 2 C_1 sin(t) + C_2 (cos(t)-sin(t))
          end{array}
          right).
          end{equation}






          share|cite|improve this answer











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            1












            $begingroup$

            You were almost done, as you can simply integrate the function $M^{-1}(t)b(t)$ over $t$ to obtain the function $C(t)$, up to constants of integration.



            You write your example in the form
            begin{equation}
            boldsymbol{dot{x}} = left( begin{array}{c}
            dot{x}\
            dot{y}
            end{array}
            right) = left( begin{array}{cc}
            1 & 1\
            -2 & -1
            end{array}
            right) left( begin{array}{c}
            x\
            y
            end{array}
            right) + left( begin{array}{c}
            -cos(t)\
            sin(t)+cos(t)
            end{array}
            right) = boldsymbol{underline{A}} boldsymbol{x} + boldsymbol{b}(t),
            end{equation}
            which has the general solution
            begin{equation}
            boldsymbol{x}(t) = boldsymbol{underline{M}}(t) int boldsymbol{underline{M}}^{-1}(t) boldsymbol{b}(t) , mathrm{d}t,
            end{equation}
            where $boldsymbol{underline{M}}(t) := e^{t boldsymbol{underline{A}}}$ is a matrix exponential (with inverse $boldsymbol{underline{M}}^{-1}(t) = e^{-t boldsymbol{underline{A}}}$). The constants of integration are contained in the indefinite integral. We obtain
            begin{equation}
            boldsymbol{underline{M}}(t) = e^{tboldsymbol{underline{A}}} = left( begin{array}{cc}
            cos(t) + sin(t) & sin(t)\
            -2sin(t) & cos(t) - sin(t)
            end{array}
            right), quad int boldsymbol{underline{M}}^{-1}(t) boldsymbol{b}(t) , mathrm{d}t = left( begin{array}{c}
            C_1 - t\
            C_2 + t
            end{array}
            right),
            end{equation}
            $C_1, C_2 in mathbb{R}$, and finally
            begin{equation}
            boldsymbol{x}(t) = left( begin{array}{c}
            x(t)\
            y(t)
            end{array}
            right) = left( begin{array}{c}
            -t cos(t) + C_1 (cos(t) + sin(t)) + C_2 sin(t)\
            t cos(t) + t sin(t) - 2 C_1 sin(t) + C_2 (cos(t)-sin(t))
            end{array}
            right).
            end{equation}






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              You were almost done, as you can simply integrate the function $M^{-1}(t)b(t)$ over $t$ to obtain the function $C(t)$, up to constants of integration.



              You write your example in the form
              begin{equation}
              boldsymbol{dot{x}} = left( begin{array}{c}
              dot{x}\
              dot{y}
              end{array}
              right) = left( begin{array}{cc}
              1 & 1\
              -2 & -1
              end{array}
              right) left( begin{array}{c}
              x\
              y
              end{array}
              right) + left( begin{array}{c}
              -cos(t)\
              sin(t)+cos(t)
              end{array}
              right) = boldsymbol{underline{A}} boldsymbol{x} + boldsymbol{b}(t),
              end{equation}
              which has the general solution
              begin{equation}
              boldsymbol{x}(t) = boldsymbol{underline{M}}(t) int boldsymbol{underline{M}}^{-1}(t) boldsymbol{b}(t) , mathrm{d}t,
              end{equation}
              where $boldsymbol{underline{M}}(t) := e^{t boldsymbol{underline{A}}}$ is a matrix exponential (with inverse $boldsymbol{underline{M}}^{-1}(t) = e^{-t boldsymbol{underline{A}}}$). The constants of integration are contained in the indefinite integral. We obtain
              begin{equation}
              boldsymbol{underline{M}}(t) = e^{tboldsymbol{underline{A}}} = left( begin{array}{cc}
              cos(t) + sin(t) & sin(t)\
              -2sin(t) & cos(t) - sin(t)
              end{array}
              right), quad int boldsymbol{underline{M}}^{-1}(t) boldsymbol{b}(t) , mathrm{d}t = left( begin{array}{c}
              C_1 - t\
              C_2 + t
              end{array}
              right),
              end{equation}
              $C_1, C_2 in mathbb{R}$, and finally
              begin{equation}
              boldsymbol{x}(t) = left( begin{array}{c}
              x(t)\
              y(t)
              end{array}
              right) = left( begin{array}{c}
              -t cos(t) + C_1 (cos(t) + sin(t)) + C_2 sin(t)\
              t cos(t) + t sin(t) - 2 C_1 sin(t) + C_2 (cos(t)-sin(t))
              end{array}
              right).
              end{equation}






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                You were almost done, as you can simply integrate the function $M^{-1}(t)b(t)$ over $t$ to obtain the function $C(t)$, up to constants of integration.



                You write your example in the form
                begin{equation}
                boldsymbol{dot{x}} = left( begin{array}{c}
                dot{x}\
                dot{y}
                end{array}
                right) = left( begin{array}{cc}
                1 & 1\
                -2 & -1
                end{array}
                right) left( begin{array}{c}
                x\
                y
                end{array}
                right) + left( begin{array}{c}
                -cos(t)\
                sin(t)+cos(t)
                end{array}
                right) = boldsymbol{underline{A}} boldsymbol{x} + boldsymbol{b}(t),
                end{equation}
                which has the general solution
                begin{equation}
                boldsymbol{x}(t) = boldsymbol{underline{M}}(t) int boldsymbol{underline{M}}^{-1}(t) boldsymbol{b}(t) , mathrm{d}t,
                end{equation}
                where $boldsymbol{underline{M}}(t) := e^{t boldsymbol{underline{A}}}$ is a matrix exponential (with inverse $boldsymbol{underline{M}}^{-1}(t) = e^{-t boldsymbol{underline{A}}}$). The constants of integration are contained in the indefinite integral. We obtain
                begin{equation}
                boldsymbol{underline{M}}(t) = e^{tboldsymbol{underline{A}}} = left( begin{array}{cc}
                cos(t) + sin(t) & sin(t)\
                -2sin(t) & cos(t) - sin(t)
                end{array}
                right), quad int boldsymbol{underline{M}}^{-1}(t) boldsymbol{b}(t) , mathrm{d}t = left( begin{array}{c}
                C_1 - t\
                C_2 + t
                end{array}
                right),
                end{equation}
                $C_1, C_2 in mathbb{R}$, and finally
                begin{equation}
                boldsymbol{x}(t) = left( begin{array}{c}
                x(t)\
                y(t)
                end{array}
                right) = left( begin{array}{c}
                -t cos(t) + C_1 (cos(t) + sin(t)) + C_2 sin(t)\
                t cos(t) + t sin(t) - 2 C_1 sin(t) + C_2 (cos(t)-sin(t))
                end{array}
                right).
                end{equation}






                share|cite|improve this answer











                $endgroup$



                You were almost done, as you can simply integrate the function $M^{-1}(t)b(t)$ over $t$ to obtain the function $C(t)$, up to constants of integration.



                You write your example in the form
                begin{equation}
                boldsymbol{dot{x}} = left( begin{array}{c}
                dot{x}\
                dot{y}
                end{array}
                right) = left( begin{array}{cc}
                1 & 1\
                -2 & -1
                end{array}
                right) left( begin{array}{c}
                x\
                y
                end{array}
                right) + left( begin{array}{c}
                -cos(t)\
                sin(t)+cos(t)
                end{array}
                right) = boldsymbol{underline{A}} boldsymbol{x} + boldsymbol{b}(t),
                end{equation}
                which has the general solution
                begin{equation}
                boldsymbol{x}(t) = boldsymbol{underline{M}}(t) int boldsymbol{underline{M}}^{-1}(t) boldsymbol{b}(t) , mathrm{d}t,
                end{equation}
                where $boldsymbol{underline{M}}(t) := e^{t boldsymbol{underline{A}}}$ is a matrix exponential (with inverse $boldsymbol{underline{M}}^{-1}(t) = e^{-t boldsymbol{underline{A}}}$). The constants of integration are contained in the indefinite integral. We obtain
                begin{equation}
                boldsymbol{underline{M}}(t) = e^{tboldsymbol{underline{A}}} = left( begin{array}{cc}
                cos(t) + sin(t) & sin(t)\
                -2sin(t) & cos(t) - sin(t)
                end{array}
                right), quad int boldsymbol{underline{M}}^{-1}(t) boldsymbol{b}(t) , mathrm{d}t = left( begin{array}{c}
                C_1 - t\
                C_2 + t
                end{array}
                right),
                end{equation}
                $C_1, C_2 in mathbb{R}$, and finally
                begin{equation}
                boldsymbol{x}(t) = left( begin{array}{c}
                x(t)\
                y(t)
                end{array}
                right) = left( begin{array}{c}
                -t cos(t) + C_1 (cos(t) + sin(t)) + C_2 sin(t)\
                t cos(t) + t sin(t) - 2 C_1 sin(t) + C_2 (cos(t)-sin(t))
                end{array}
                right).
                end{equation}







                share|cite|improve this answer














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                edited Jan 15 at 5:54

























                answered Jan 15 at 5:45









                ChristophChristoph

                58116




                58116






























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