System of differential equations with t on one side











up vote
2
down vote

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I got this system
$$
left{
begin{array}{c}
dx/dt+6x-5y-t=0 \
dy/dt+5x-4y-1=0 \
end{array}
right.
$$



which leads to:
$$
left{
begin{array}{c}
(D+6)x-5y-t=0 \
(D-4)y+5x-1=0 \
end{array}
right.
$$



I multiplied top one by $(D-4)$ and bottom one by $5$ to cancel out ys.
In result I got
$$
(D+1)^2x=(D-4)t+5
$$
Using repeated root formula, $Xc=c_1e^{-t}+c_2te^{-t}$



Now my problem starts. What should I do with $Dt-4t+5$. All examples in book have something like $e^t$ or $t^2$, which is easy to take derivative of. In my case, would it just be $-4t + 5 + C$?



If so, would trial particular solution be $Xp=At+B$? Do I just ignore "C"?



Thanks!






differential-equations systems-of-equations







share|cite|improve this question


























    up vote
    2
    down vote

    favorite












    I got this system
    $$
    left{
    begin{array}{c}
    dx/dt+6x-5y-t=0 \
    dy/dt+5x-4y-1=0 \
    end{array}
    right.
    $$



    which leads to:
    $$
    left{
    begin{array}{c}
    (D+6)x-5y-t=0 \
    (D-4)y+5x-1=0 \
    end{array}
    right.
    $$



    I multiplied top one by $(D-4)$ and bottom one by $5$ to cancel out ys.
    In result I got
    $$
    (D+1)^2x=(D-4)t+5
    $$
    Using repeated root formula, $Xc=c_1e^{-t}+c_2te^{-t}$



    Now my problem starts. What should I do with $Dt-4t+5$. All examples in book have something like $e^t$ or $t^2$, which is easy to take derivative of. In my case, would it just be $-4t + 5 + C$?



    If so, would trial particular solution be $Xp=At+B$? Do I just ignore "C"?



    Thanks!






    differential-equations systems-of-equations







    share|cite|improve this question
























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I got this system
      $$
      left{
      begin{array}{c}
      dx/dt+6x-5y-t=0 \
      dy/dt+5x-4y-1=0 \
      end{array}
      right.
      $$



      which leads to:
      $$
      left{
      begin{array}{c}
      (D+6)x-5y-t=0 \
      (D-4)y+5x-1=0 \
      end{array}
      right.
      $$



      I multiplied top one by $(D-4)$ and bottom one by $5$ to cancel out ys.
      In result I got
      $$
      (D+1)^2x=(D-4)t+5
      $$
      Using repeated root formula, $Xc=c_1e^{-t}+c_2te^{-t}$



      Now my problem starts. What should I do with $Dt-4t+5$. All examples in book have something like $e^t$ or $t^2$, which is easy to take derivative of. In my case, would it just be $-4t + 5 + C$?



      If so, would trial particular solution be $Xp=At+B$? Do I just ignore "C"?



      Thanks!






      differential-equations systems-of-equations







      share|cite|improve this question













      I got this system
      $$
      left{
      begin{array}{c}
      dx/dt+6x-5y-t=0 \
      dy/dt+5x-4y-1=0 \
      end{array}
      right.
      $$



      which leads to:
      $$
      left{
      begin{array}{c}
      (D+6)x-5y-t=0 \
      (D-4)y+5x-1=0 \
      end{array}
      right.
      $$



      I multiplied top one by $(D-4)$ and bottom one by $5$ to cancel out ys.
      In result I got
      $$
      (D+1)^2x=(D-4)t+5
      $$
      Using repeated root formula, $Xc=c_1e^{-t}+c_2te^{-t}$



      Now my problem starts. What should I do with $Dt-4t+5$. All examples in book have something like $e^t$ or $t^2$, which is easy to take derivative of. In my case, would it just be $-4t + 5 + C$?



      If so, would trial particular solution be $Xp=At+B$? Do I just ignore "C"?



      Thanks!






      differential-equations systems-of-equations




      differential-equations systems-of-equations






      share|cite|improve this question













      share|cite|improve this question











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      asked Nov 13 at 0:07









      Kolcek2384

      303




      303






















          1 Answer
          1






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          up vote
          1
          down vote



          accepted










          $$(D+1)^2x=(D-4)t+5$$
          $$ (D+1)^2x=1-4t+5$$
          So we have
          $$implies (D+1)^2x=-4t+6$$



          Try this as particular solution
          $$x_h=At+B$$
          $$(D+1)^2(At+B)=-4t+6$$
          $$implies At +2A+B=-4t+6$$
          Solve the system to get A and B



          Note that the constant C appears when you integrate not when you differentiate...






          share|cite|improve this answer



















          • 1




            @Kolcek2384 $$D(t)=1$$ $$D(t^2)=2t$$ $$D(e^t)=e^t$$ I added 1
            – Isham
            Nov 13 at 0:36








          • 1




            You sure it's 13 ??? For A you are right and the general solution is ok @Kolcek2384
            – Isham
            Nov 13 at 0:37






          • 1




            It lookks good to me ..@Kolcek2384
            – Isham
            Nov 13 at 0:41






          • 1




            Seems like easiest way to solve for y would be to use first equation $Dx+6x-5y-t=0$. So $y=frac{Dx+6x-t}{5}$. $Dx=-c_1e^{-t}+c_2e^{-t}-c_2te^{-t}-4$. In this case $$y=c_1e^{-t}+c_2te^{-t}+frac{1}{5}c_2e^{-t}-5t+16 $$
            – Kolcek2384
            Nov 13 at 0:56








          • 1




            Yes thats the best way....@Kolcek2384
            – Isham
            Nov 13 at 0:57











          Your Answer





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          up vote
          1
          down vote



          accepted










          $$(D+1)^2x=(D-4)t+5$$
          $$ (D+1)^2x=1-4t+5$$
          So we have
          $$implies (D+1)^2x=-4t+6$$



          Try this as particular solution
          $$x_h=At+B$$
          $$(D+1)^2(At+B)=-4t+6$$
          $$implies At +2A+B=-4t+6$$
          Solve the system to get A and B



          Note that the constant C appears when you integrate not when you differentiate...






          share|cite|improve this answer



















          • 1




            @Kolcek2384 $$D(t)=1$$ $$D(t^2)=2t$$ $$D(e^t)=e^t$$ I added 1
            – Isham
            Nov 13 at 0:36








          • 1




            You sure it's 13 ??? For A you are right and the general solution is ok @Kolcek2384
            – Isham
            Nov 13 at 0:37






          • 1




            It lookks good to me ..@Kolcek2384
            – Isham
            Nov 13 at 0:41






          • 1




            Seems like easiest way to solve for y would be to use first equation $Dx+6x-5y-t=0$. So $y=frac{Dx+6x-t}{5}$. $Dx=-c_1e^{-t}+c_2e^{-t}-c_2te^{-t}-4$. In this case $$y=c_1e^{-t}+c_2te^{-t}+frac{1}{5}c_2e^{-t}-5t+16 $$
            – Kolcek2384
            Nov 13 at 0:56








          • 1




            Yes thats the best way....@Kolcek2384
            – Isham
            Nov 13 at 0:57















          up vote
          1
          down vote



          accepted










          $$(D+1)^2x=(D-4)t+5$$
          $$ (D+1)^2x=1-4t+5$$
          So we have
          $$implies (D+1)^2x=-4t+6$$



          Try this as particular solution
          $$x_h=At+B$$
          $$(D+1)^2(At+B)=-4t+6$$
          $$implies At +2A+B=-4t+6$$
          Solve the system to get A and B



          Note that the constant C appears when you integrate not when you differentiate...






          share|cite|improve this answer



















          • 1




            @Kolcek2384 $$D(t)=1$$ $$D(t^2)=2t$$ $$D(e^t)=e^t$$ I added 1
            – Isham
            Nov 13 at 0:36








          • 1




            You sure it's 13 ??? For A you are right and the general solution is ok @Kolcek2384
            – Isham
            Nov 13 at 0:37






          • 1




            It lookks good to me ..@Kolcek2384
            – Isham
            Nov 13 at 0:41






          • 1




            Seems like easiest way to solve for y would be to use first equation $Dx+6x-5y-t=0$. So $y=frac{Dx+6x-t}{5}$. $Dx=-c_1e^{-t}+c_2e^{-t}-c_2te^{-t}-4$. In this case $$y=c_1e^{-t}+c_2te^{-t}+frac{1}{5}c_2e^{-t}-5t+16 $$
            – Kolcek2384
            Nov 13 at 0:56








          • 1




            Yes thats the best way....@Kolcek2384
            – Isham
            Nov 13 at 0:57













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          $$(D+1)^2x=(D-4)t+5$$
          $$ (D+1)^2x=1-4t+5$$
          So we have
          $$implies (D+1)^2x=-4t+6$$



          Try this as particular solution
          $$x_h=At+B$$
          $$(D+1)^2(At+B)=-4t+6$$
          $$implies At +2A+B=-4t+6$$
          Solve the system to get A and B



          Note that the constant C appears when you integrate not when you differentiate...






          share|cite|improve this answer














          $$(D+1)^2x=(D-4)t+5$$
          $$ (D+1)^2x=1-4t+5$$
          So we have
          $$implies (D+1)^2x=-4t+6$$



          Try this as particular solution
          $$x_h=At+B$$
          $$(D+1)^2(At+B)=-4t+6$$
          $$implies At +2A+B=-4t+6$$
          Solve the system to get A and B



          Note that the constant C appears when you integrate not when you differentiate...







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 13 at 0:28

























          answered Nov 13 at 0:22









          Isham

          12.6k3929




          12.6k3929








          • 1




            @Kolcek2384 $$D(t)=1$$ $$D(t^2)=2t$$ $$D(e^t)=e^t$$ I added 1
            – Isham
            Nov 13 at 0:36








          • 1




            You sure it's 13 ??? For A you are right and the general solution is ok @Kolcek2384
            – Isham
            Nov 13 at 0:37






          • 1




            It lookks good to me ..@Kolcek2384
            – Isham
            Nov 13 at 0:41






          • 1




            Seems like easiest way to solve for y would be to use first equation $Dx+6x-5y-t=0$. So $y=frac{Dx+6x-t}{5}$. $Dx=-c_1e^{-t}+c_2e^{-t}-c_2te^{-t}-4$. In this case $$y=c_1e^{-t}+c_2te^{-t}+frac{1}{5}c_2e^{-t}-5t+16 $$
            – Kolcek2384
            Nov 13 at 0:56








          • 1




            Yes thats the best way....@Kolcek2384
            – Isham
            Nov 13 at 0:57














          • 1




            @Kolcek2384 $$D(t)=1$$ $$D(t^2)=2t$$ $$D(e^t)=e^t$$ I added 1
            – Isham
            Nov 13 at 0:36








          • 1




            You sure it's 13 ??? For A you are right and the general solution is ok @Kolcek2384
            – Isham
            Nov 13 at 0:37






          • 1




            It lookks good to me ..@Kolcek2384
            – Isham
            Nov 13 at 0:41






          • 1




            Seems like easiest way to solve for y would be to use first equation $Dx+6x-5y-t=0$. So $y=frac{Dx+6x-t}{5}$. $Dx=-c_1e^{-t}+c_2e^{-t}-c_2te^{-t}-4$. In this case $$y=c_1e^{-t}+c_2te^{-t}+frac{1}{5}c_2e^{-t}-5t+16 $$
            – Kolcek2384
            Nov 13 at 0:56








          • 1




            Yes thats the best way....@Kolcek2384
            – Isham
            Nov 13 at 0:57








          1




          1




          @Kolcek2384 $$D(t)=1$$ $$D(t^2)=2t$$ $$D(e^t)=e^t$$ I added 1
          – Isham
          Nov 13 at 0:36






          @Kolcek2384 $$D(t)=1$$ $$D(t^2)=2t$$ $$D(e^t)=e^t$$ I added 1
          – Isham
          Nov 13 at 0:36






          1




          1




          You sure it's 13 ??? For A you are right and the general solution is ok @Kolcek2384
          – Isham
          Nov 13 at 0:37




          You sure it's 13 ??? For A you are right and the general solution is ok @Kolcek2384
          – Isham
          Nov 13 at 0:37




          1




          1




          It lookks good to me ..@Kolcek2384
          – Isham
          Nov 13 at 0:41




          It lookks good to me ..@Kolcek2384
          – Isham
          Nov 13 at 0:41




          1




          1




          Seems like easiest way to solve for y would be to use first equation $Dx+6x-5y-t=0$. So $y=frac{Dx+6x-t}{5}$. $Dx=-c_1e^{-t}+c_2e^{-t}-c_2te^{-t}-4$. In this case $$y=c_1e^{-t}+c_2te^{-t}+frac{1}{5}c_2e^{-t}-5t+16 $$
          – Kolcek2384
          Nov 13 at 0:56






          Seems like easiest way to solve for y would be to use first equation $Dx+6x-5y-t=0$. So $y=frac{Dx+6x-t}{5}$. $Dx=-c_1e^{-t}+c_2e^{-t}-c_2te^{-t}-4$. In this case $$y=c_1e^{-t}+c_2te^{-t}+frac{1}{5}c_2e^{-t}-5t+16 $$
          – Kolcek2384
          Nov 13 at 0:56






          1




          1




          Yes thats the best way....@Kolcek2384
          – Isham
          Nov 13 at 0:57




          Yes thats the best way....@Kolcek2384
          – Isham
          Nov 13 at 0:57


















           

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