Help solving a system of differential equations












0












$begingroup$


Consider the third order linear differential equation:



$x'''- 2x''- 4x'+ 8x = 0.$



Given that $x(0) = 4$, $x'(0) = 16$, and $x''(0) = 16$, find the general solution to the corresponding first-order system.



I know that if I set $x_1 = x$, $x_2 = x_1'$, and $x_3 = x_2'$, I can generate the following system of equations:



$x_1' = x_2$,



$x_2' = x_3$,



$x_3' = -8x_1+4x_2+2x_3$.



I have got 2 (multiplicity of 2) and -2 (multiplicity of 1) for eigenvalues, but I'm not sure how to finish from here. Help would be appreciated.






linear-algebra ordinary-differential-equations







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  • $begingroup$
    Yes, thanks for pointing that out, I wrote a typo.
    $endgroup$
    – realgneel
    Nov 29 '18 at 9:48
















0












$begingroup$


Consider the third order linear differential equation:



$x'''- 2x''- 4x'+ 8x = 0.$



Given that $x(0) = 4$, $x'(0) = 16$, and $x''(0) = 16$, find the general solution to the corresponding first-order system.



I know that if I set $x_1 = x$, $x_2 = x_1'$, and $x_3 = x_2'$, I can generate the following system of equations:



$x_1' = x_2$,



$x_2' = x_3$,



$x_3' = -8x_1+4x_2+2x_3$.



I have got 2 (multiplicity of 2) and -2 (multiplicity of 1) for eigenvalues, but I'm not sure how to finish from here. Help would be appreciated.






linear-algebra ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, thanks for pointing that out, I wrote a typo.
    $endgroup$
    – realgneel
    Nov 29 '18 at 9:48














0












0








0





$begingroup$


Consider the third order linear differential equation:



$x'''- 2x''- 4x'+ 8x = 0.$



Given that $x(0) = 4$, $x'(0) = 16$, and $x''(0) = 16$, find the general solution to the corresponding first-order system.



I know that if I set $x_1 = x$, $x_2 = x_1'$, and $x_3 = x_2'$, I can generate the following system of equations:



$x_1' = x_2$,



$x_2' = x_3$,



$x_3' = -8x_1+4x_2+2x_3$.



I have got 2 (multiplicity of 2) and -2 (multiplicity of 1) for eigenvalues, but I'm not sure how to finish from here. Help would be appreciated.






linear-algebra ordinary-differential-equations







share|cite|improve this question











$endgroup$




Consider the third order linear differential equation:



$x'''- 2x''- 4x'+ 8x = 0.$



Given that $x(0) = 4$, $x'(0) = 16$, and $x''(0) = 16$, find the general solution to the corresponding first-order system.



I know that if I set $x_1 = x$, $x_2 = x_1'$, and $x_3 = x_2'$, I can generate the following system of equations:



$x_1' = x_2$,



$x_2' = x_3$,



$x_3' = -8x_1+4x_2+2x_3$.



I have got 2 (multiplicity of 2) and -2 (multiplicity of 1) for eigenvalues, but I'm not sure how to finish from here. Help would be appreciated.






linear-algebra ordinary-differential-equations




linear-algebra ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 9:47







realgneel

















asked Nov 29 '18 at 8:31









realgneelrealgneel

32




32












  • $begingroup$
    Yes, thanks for pointing that out, I wrote a typo.
    $endgroup$
    – realgneel
    Nov 29 '18 at 9:48


















  • $begingroup$
    Yes, thanks for pointing that out, I wrote a typo.
    $endgroup$
    – realgneel
    Nov 29 '18 at 9:48
















$begingroup$
Yes, thanks for pointing that out, I wrote a typo.
$endgroup$
– realgneel
Nov 29 '18 at 9:48




$begingroup$
Yes, thanks for pointing that out, I wrote a typo.
$endgroup$
– realgneel
Nov 29 '18 at 9:48










2 Answers
2






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1












$begingroup$

I think you have an error in your equation. Looking at your expression for $x_3$ it should be:



$$x'''- 2x''- 4x'+ 8x = 0$$



...and the characteristic equation is:



$$r^3-2r^2-4r+8=0$$



$$(r^3+8)-(2r^2+4r)=0$$



$$(r+2)(r^2-2r+4) -2r(r+2)=0$$



$$(r+2)(r^2-4r+4)=0$$



$$(r+2)(r-2)^2=0$$



$$r_1=-2, quad r_{2,3}=2$$



So the solution to your equation is:



$$x(t)=C_1e^{-2t}+C_2e^{2t}+C_3te^{2t}$$



Constants can be evaluated from your initial conditions. You should be able to proceed from here.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    I will assume the equation is meant to be $x'''-2x''-4x'+8x=0$, in which case its characteristic polynomial is indeed $(r-2)^2(r+2)$ and the rest of the post makes sense.



    The general solution is



    $x(t)=Ae^{-2t} +(Bt+C)e^{2t}$



    for constants $A, B, C$. So we have



    $x(0) = A+C = 4$



    $x'(0) = -2A + B +2C = 16$



    $x''(0) = 4A + 4B + 4C = 16$



    Now solve these simultaneous equations in $A,B,C$ to find the specific solution that meets your boundary conditions.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

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      1












      $begingroup$

      I think you have an error in your equation. Looking at your expression for $x_3$ it should be:



      $$x'''- 2x''- 4x'+ 8x = 0$$



      ...and the characteristic equation is:



      $$r^3-2r^2-4r+8=0$$



      $$(r^3+8)-(2r^2+4r)=0$$



      $$(r+2)(r^2-2r+4) -2r(r+2)=0$$



      $$(r+2)(r^2-4r+4)=0$$



      $$(r+2)(r-2)^2=0$$



      $$r_1=-2, quad r_{2,3}=2$$



      So the solution to your equation is:



      $$x(t)=C_1e^{-2t}+C_2e^{2t}+C_3te^{2t}$$



      Constants can be evaluated from your initial conditions. You should be able to proceed from here.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        I think you have an error in your equation. Looking at your expression for $x_3$ it should be:



        $$x'''- 2x''- 4x'+ 8x = 0$$



        ...and the characteristic equation is:



        $$r^3-2r^2-4r+8=0$$



        $$(r^3+8)-(2r^2+4r)=0$$



        $$(r+2)(r^2-2r+4) -2r(r+2)=0$$



        $$(r+2)(r^2-4r+4)=0$$



        $$(r+2)(r-2)^2=0$$



        $$r_1=-2, quad r_{2,3}=2$$



        So the solution to your equation is:



        $$x(t)=C_1e^{-2t}+C_2e^{2t}+C_3te^{2t}$$



        Constants can be evaluated from your initial conditions. You should be able to proceed from here.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          I think you have an error in your equation. Looking at your expression for $x_3$ it should be:



          $$x'''- 2x''- 4x'+ 8x = 0$$



          ...and the characteristic equation is:



          $$r^3-2r^2-4r+8=0$$



          $$(r^3+8)-(2r^2+4r)=0$$



          $$(r+2)(r^2-2r+4) -2r(r+2)=0$$



          $$(r+2)(r^2-4r+4)=0$$



          $$(r+2)(r-2)^2=0$$



          $$r_1=-2, quad r_{2,3}=2$$



          So the solution to your equation is:



          $$x(t)=C_1e^{-2t}+C_2e^{2t}+C_3te^{2t}$$



          Constants can be evaluated from your initial conditions. You should be able to proceed from here.






          share|cite|improve this answer









          $endgroup$



          I think you have an error in your equation. Looking at your expression for $x_3$ it should be:



          $$x'''- 2x''- 4x'+ 8x = 0$$



          ...and the characteristic equation is:



          $$r^3-2r^2-4r+8=0$$



          $$(r^3+8)-(2r^2+4r)=0$$



          $$(r+2)(r^2-2r+4) -2r(r+2)=0$$



          $$(r+2)(r^2-4r+4)=0$$



          $$(r+2)(r-2)^2=0$$



          $$r_1=-2, quad r_{2,3}=2$$



          So the solution to your equation is:



          $$x(t)=C_1e^{-2t}+C_2e^{2t}+C_3te^{2t}$$



          Constants can be evaluated from your initial conditions. You should be able to proceed from here.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 9:00









          OldboyOldboy

          7,9301935




          7,9301935























              0












              $begingroup$

              I will assume the equation is meant to be $x'''-2x''-4x'+8x=0$, in which case its characteristic polynomial is indeed $(r-2)^2(r+2)$ and the rest of the post makes sense.



              The general solution is



              $x(t)=Ae^{-2t} +(Bt+C)e^{2t}$



              for constants $A, B, C$. So we have



              $x(0) = A+C = 4$



              $x'(0) = -2A + B +2C = 16$



              $x''(0) = 4A + 4B + 4C = 16$



              Now solve these simultaneous equations in $A,B,C$ to find the specific solution that meets your boundary conditions.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                I will assume the equation is meant to be $x'''-2x''-4x'+8x=0$, in which case its characteristic polynomial is indeed $(r-2)^2(r+2)$ and the rest of the post makes sense.



                The general solution is



                $x(t)=Ae^{-2t} +(Bt+C)e^{2t}$



                for constants $A, B, C$. So we have



                $x(0) = A+C = 4$



                $x'(0) = -2A + B +2C = 16$



                $x''(0) = 4A + 4B + 4C = 16$



                Now solve these simultaneous equations in $A,B,C$ to find the specific solution that meets your boundary conditions.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I will assume the equation is meant to be $x'''-2x''-4x'+8x=0$, in which case its characteristic polynomial is indeed $(r-2)^2(r+2)$ and the rest of the post makes sense.



                  The general solution is



                  $x(t)=Ae^{-2t} +(Bt+C)e^{2t}$



                  for constants $A, B, C$. So we have



                  $x(0) = A+C = 4$



                  $x'(0) = -2A + B +2C = 16$



                  $x''(0) = 4A + 4B + 4C = 16$



                  Now solve these simultaneous equations in $A,B,C$ to find the specific solution that meets your boundary conditions.






                  share|cite|improve this answer









                  $endgroup$



                  I will assume the equation is meant to be $x'''-2x''-4x'+8x=0$, in which case its characteristic polynomial is indeed $(r-2)^2(r+2)$ and the rest of the post makes sense.



                  The general solution is



                  $x(t)=Ae^{-2t} +(Bt+C)e^{2t}$



                  for constants $A, B, C$. So we have



                  $x(0) = A+C = 4$



                  $x'(0) = -2A + B +2C = 16$



                  $x''(0) = 4A + 4B + 4C = 16$



                  Now solve these simultaneous equations in $A,B,C$ to find the specific solution that meets your boundary conditions.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 29 '18 at 9:08









                  gandalf61gandalf61

                  8,536725




                  8,536725






























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