I need help finding the general solution to the differential equation $y''(t)+7y'(t)=-14$











up vote
2
down vote

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What I've tried:



I have the inhomogeneous differential equation:



$$y''(t)+7y'(t)=-14$$



I find the particular solution to be on the form $$kt$$



by inserting the particular solution in the equation



$$(kt)''+7(kt)'=-14$$



and isolating for k, I get that:



$$k=-2$$



and therefore the particular solution is



$$y(t)-2t$$



I also need the general solution for the homogenous equation



$$y''(t)+7y'(t)=0$$



by finding the roots of the characteristic polynomial



$$z^2+7z=z(z+7)=0$$
$$z_1=0$$
$$z_2=-7$$



I get the general solution:



$$c_1e^{0t}+c_2e^{-7t}=c_1+c_2e^{-7t}$$



Now, according to my textbook, the general solution of an inhomogeneous differential equation is given by



$$y(t)=y_p(t)+y_{hom}(t)$$



Where $y_p(t)$ is the particular solution and $y_{hom}(t)$ is the general solution to the homogenous equation. Therefore I get the general solution to be



$$y(t)=c_1+c_2e^{-7t}-2t$$



This is not consistent with Maple's result however



enter image description here



Can anyone see where I've gone wrong?










share|cite|improve this question


























    up vote
    2
    down vote

    favorite












    What I've tried:



    I have the inhomogeneous differential equation:



    $$y''(t)+7y'(t)=-14$$



    I find the particular solution to be on the form $$kt$$



    by inserting the particular solution in the equation



    $$(kt)''+7(kt)'=-14$$



    and isolating for k, I get that:



    $$k=-2$$



    and therefore the particular solution is



    $$y(t)-2t$$



    I also need the general solution for the homogenous equation



    $$y''(t)+7y'(t)=0$$



    by finding the roots of the characteristic polynomial



    $$z^2+7z=z(z+7)=0$$
    $$z_1=0$$
    $$z_2=-7$$



    I get the general solution:



    $$c_1e^{0t}+c_2e^{-7t}=c_1+c_2e^{-7t}$$



    Now, according to my textbook, the general solution of an inhomogeneous differential equation is given by



    $$y(t)=y_p(t)+y_{hom}(t)$$



    Where $y_p(t)$ is the particular solution and $y_{hom}(t)$ is the general solution to the homogenous equation. Therefore I get the general solution to be



    $$y(t)=c_1+c_2e^{-7t}-2t$$



    This is not consistent with Maple's result however



    enter image description here



    Can anyone see where I've gone wrong?










    share|cite|improve this question
























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      What I've tried:



      I have the inhomogeneous differential equation:



      $$y''(t)+7y'(t)=-14$$



      I find the particular solution to be on the form $$kt$$



      by inserting the particular solution in the equation



      $$(kt)''+7(kt)'=-14$$



      and isolating for k, I get that:



      $$k=-2$$



      and therefore the particular solution is



      $$y(t)-2t$$



      I also need the general solution for the homogenous equation



      $$y''(t)+7y'(t)=0$$



      by finding the roots of the characteristic polynomial



      $$z^2+7z=z(z+7)=0$$
      $$z_1=0$$
      $$z_2=-7$$



      I get the general solution:



      $$c_1e^{0t}+c_2e^{-7t}=c_1+c_2e^{-7t}$$



      Now, according to my textbook, the general solution of an inhomogeneous differential equation is given by



      $$y(t)=y_p(t)+y_{hom}(t)$$



      Where $y_p(t)$ is the particular solution and $y_{hom}(t)$ is the general solution to the homogenous equation. Therefore I get the general solution to be



      $$y(t)=c_1+c_2e^{-7t}-2t$$



      This is not consistent with Maple's result however



      enter image description here



      Can anyone see where I've gone wrong?










      share|cite|improve this question













      What I've tried:



      I have the inhomogeneous differential equation:



      $$y''(t)+7y'(t)=-14$$



      I find the particular solution to be on the form $$kt$$



      by inserting the particular solution in the equation



      $$(kt)''+7(kt)'=-14$$



      and isolating for k, I get that:



      $$k=-2$$



      and therefore the particular solution is



      $$y(t)-2t$$



      I also need the general solution for the homogenous equation



      $$y''(t)+7y'(t)=0$$



      by finding the roots of the characteristic polynomial



      $$z^2+7z=z(z+7)=0$$
      $$z_1=0$$
      $$z_2=-7$$



      I get the general solution:



      $$c_1e^{0t}+c_2e^{-7t}=c_1+c_2e^{-7t}$$



      Now, according to my textbook, the general solution of an inhomogeneous differential equation is given by



      $$y(t)=y_p(t)+y_{hom}(t)$$



      Where $y_p(t)$ is the particular solution and $y_{hom}(t)$ is the general solution to the homogenous equation. Therefore I get the general solution to be



      $$y(t)=c_1+c_2e^{-7t}-2t$$



      This is not consistent with Maple's result however



      enter image description here



      Can anyone see where I've gone wrong?







      differential-equations






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      share|cite|improve this question











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      asked Nov 19 at 21:27









      Boris Grunwald

      1487




      1487






















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          You have not gone wrong anywhere! It is just a question concerning the choice of the arbitrary constants $c_1,c_2$ and $C_1,C_2$ respectively. Therefore lets just take a brief look at the two given solutions



          $$begin{align}
          y_1(t)&=c_1+c_2e^{-7t}-2t\
          y_2(t)&=C_2+frac{-C_1}{7}e^{-7t}-2t
          end{align}$$



          We can agree one the fact that the particular solution $-2t$ is the same in both $y_1(t)$ and $y_2(t)$. Therefore we do not have to worry about this term further. Now we can argue that $c_1=C_2$ since these are both arbitrary constants which have to be defined later one. Following a similiar agrumentation we can moreover say that $c_2=frac{-C_1}{7}$ hence there is no restriction regarding their value.



          Therefore Maple aswell as you by yourself provided the exact same solution beside the choice of some arbitrary constants.






          share|cite|improve this answer





















          • Thanks, that makes sense!
            – Boris Grunwald
            Nov 19 at 21:46


















          up vote
          1
          down vote













          You went wrong when you thought what Maple wrote is different from your solution in any significant way.



          Maple has swapped the roles of $c_1$ and $c_2$ compared to you. And Maple's $c_1$ is seven times larger than your $c_2$ and has opposite sign, but since the constants are arbitrary anyways this doesn't matter. So you and Maple describe the exact same collection of functions.






          share|cite|improve this answer




























            up vote
            0
            down vote













            As others said, your solution is right (only the expressions of the constant differ). Here is how I would solve.



            First notice that $y$ is missing, and solve for $z:=y'$:



            $$z'+7z=-14.$$



            An obvious particular solution is the constant $z=-2$. Then with the characteristic polynomial $p+7$, you can immediately write the general solution,



            $$z=ce^{-7t}-2.$$



            Now to get $y$, you integrate once, and obtain



            $$y=c_0e^{-7t}-2t+c_1.$$






            share|cite|improve this answer





















              Your Answer





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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              1
              down vote



              accepted










              You have not gone wrong anywhere! It is just a question concerning the choice of the arbitrary constants $c_1,c_2$ and $C_1,C_2$ respectively. Therefore lets just take a brief look at the two given solutions



              $$begin{align}
              y_1(t)&=c_1+c_2e^{-7t}-2t\
              y_2(t)&=C_2+frac{-C_1}{7}e^{-7t}-2t
              end{align}$$



              We can agree one the fact that the particular solution $-2t$ is the same in both $y_1(t)$ and $y_2(t)$. Therefore we do not have to worry about this term further. Now we can argue that $c_1=C_2$ since these are both arbitrary constants which have to be defined later one. Following a similiar agrumentation we can moreover say that $c_2=frac{-C_1}{7}$ hence there is no restriction regarding their value.



              Therefore Maple aswell as you by yourself provided the exact same solution beside the choice of some arbitrary constants.






              share|cite|improve this answer





















              • Thanks, that makes sense!
                – Boris Grunwald
                Nov 19 at 21:46















              up vote
              1
              down vote



              accepted










              You have not gone wrong anywhere! It is just a question concerning the choice of the arbitrary constants $c_1,c_2$ and $C_1,C_2$ respectively. Therefore lets just take a brief look at the two given solutions



              $$begin{align}
              y_1(t)&=c_1+c_2e^{-7t}-2t\
              y_2(t)&=C_2+frac{-C_1}{7}e^{-7t}-2t
              end{align}$$



              We can agree one the fact that the particular solution $-2t$ is the same in both $y_1(t)$ and $y_2(t)$. Therefore we do not have to worry about this term further. Now we can argue that $c_1=C_2$ since these are both arbitrary constants which have to be defined later one. Following a similiar agrumentation we can moreover say that $c_2=frac{-C_1}{7}$ hence there is no restriction regarding their value.



              Therefore Maple aswell as you by yourself provided the exact same solution beside the choice of some arbitrary constants.






              share|cite|improve this answer





















              • Thanks, that makes sense!
                – Boris Grunwald
                Nov 19 at 21:46













              up vote
              1
              down vote



              accepted







              up vote
              1
              down vote



              accepted






              You have not gone wrong anywhere! It is just a question concerning the choice of the arbitrary constants $c_1,c_2$ and $C_1,C_2$ respectively. Therefore lets just take a brief look at the two given solutions



              $$begin{align}
              y_1(t)&=c_1+c_2e^{-7t}-2t\
              y_2(t)&=C_2+frac{-C_1}{7}e^{-7t}-2t
              end{align}$$



              We can agree one the fact that the particular solution $-2t$ is the same in both $y_1(t)$ and $y_2(t)$. Therefore we do not have to worry about this term further. Now we can argue that $c_1=C_2$ since these are both arbitrary constants which have to be defined later one. Following a similiar agrumentation we can moreover say that $c_2=frac{-C_1}{7}$ hence there is no restriction regarding their value.



              Therefore Maple aswell as you by yourself provided the exact same solution beside the choice of some arbitrary constants.






              share|cite|improve this answer












              You have not gone wrong anywhere! It is just a question concerning the choice of the arbitrary constants $c_1,c_2$ and $C_1,C_2$ respectively. Therefore lets just take a brief look at the two given solutions



              $$begin{align}
              y_1(t)&=c_1+c_2e^{-7t}-2t\
              y_2(t)&=C_2+frac{-C_1}{7}e^{-7t}-2t
              end{align}$$



              We can agree one the fact that the particular solution $-2t$ is the same in both $y_1(t)$ and $y_2(t)$. Therefore we do not have to worry about this term further. Now we can argue that $c_1=C_2$ since these are both arbitrary constants which have to be defined later one. Following a similiar agrumentation we can moreover say that $c_2=frac{-C_1}{7}$ hence there is no restriction regarding their value.



              Therefore Maple aswell as you by yourself provided the exact same solution beside the choice of some arbitrary constants.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 19 at 21:36









              mrtaurho

              2,7391827




              2,7391827












              • Thanks, that makes sense!
                – Boris Grunwald
                Nov 19 at 21:46


















              • Thanks, that makes sense!
                – Boris Grunwald
                Nov 19 at 21:46
















              Thanks, that makes sense!
              – Boris Grunwald
              Nov 19 at 21:46




              Thanks, that makes sense!
              – Boris Grunwald
              Nov 19 at 21:46










              up vote
              1
              down vote













              You went wrong when you thought what Maple wrote is different from your solution in any significant way.



              Maple has swapped the roles of $c_1$ and $c_2$ compared to you. And Maple's $c_1$ is seven times larger than your $c_2$ and has opposite sign, but since the constants are arbitrary anyways this doesn't matter. So you and Maple describe the exact same collection of functions.






              share|cite|improve this answer

























                up vote
                1
                down vote













                You went wrong when you thought what Maple wrote is different from your solution in any significant way.



                Maple has swapped the roles of $c_1$ and $c_2$ compared to you. And Maple's $c_1$ is seven times larger than your $c_2$ and has opposite sign, but since the constants are arbitrary anyways this doesn't matter. So you and Maple describe the exact same collection of functions.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  You went wrong when you thought what Maple wrote is different from your solution in any significant way.



                  Maple has swapped the roles of $c_1$ and $c_2$ compared to you. And Maple's $c_1$ is seven times larger than your $c_2$ and has opposite sign, but since the constants are arbitrary anyways this doesn't matter. So you and Maple describe the exact same collection of functions.






                  share|cite|improve this answer












                  You went wrong when you thought what Maple wrote is different from your solution in any significant way.



                  Maple has swapped the roles of $c_1$ and $c_2$ compared to you. And Maple's $c_1$ is seven times larger than your $c_2$ and has opposite sign, but since the constants are arbitrary anyways this doesn't matter. So you and Maple describe the exact same collection of functions.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 19 at 21:36









                  Arthur

                  109k7103186




                  109k7103186






















                      up vote
                      0
                      down vote













                      As others said, your solution is right (only the expressions of the constant differ). Here is how I would solve.



                      First notice that $y$ is missing, and solve for $z:=y'$:



                      $$z'+7z=-14.$$



                      An obvious particular solution is the constant $z=-2$. Then with the characteristic polynomial $p+7$, you can immediately write the general solution,



                      $$z=ce^{-7t}-2.$$



                      Now to get $y$, you integrate once, and obtain



                      $$y=c_0e^{-7t}-2t+c_1.$$






                      share|cite|improve this answer

























                        up vote
                        0
                        down vote













                        As others said, your solution is right (only the expressions of the constant differ). Here is how I would solve.



                        First notice that $y$ is missing, and solve for $z:=y'$:



                        $$z'+7z=-14.$$



                        An obvious particular solution is the constant $z=-2$. Then with the characteristic polynomial $p+7$, you can immediately write the general solution,



                        $$z=ce^{-7t}-2.$$



                        Now to get $y$, you integrate once, and obtain



                        $$y=c_0e^{-7t}-2t+c_1.$$






                        share|cite|improve this answer























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          As others said, your solution is right (only the expressions of the constant differ). Here is how I would solve.



                          First notice that $y$ is missing, and solve for $z:=y'$:



                          $$z'+7z=-14.$$



                          An obvious particular solution is the constant $z=-2$. Then with the characteristic polynomial $p+7$, you can immediately write the general solution,



                          $$z=ce^{-7t}-2.$$



                          Now to get $y$, you integrate once, and obtain



                          $$y=c_0e^{-7t}-2t+c_1.$$






                          share|cite|improve this answer












                          As others said, your solution is right (only the expressions of the constant differ). Here is how I would solve.



                          First notice that $y$ is missing, and solve for $z:=y'$:



                          $$z'+7z=-14.$$



                          An obvious particular solution is the constant $z=-2$. Then with the characteristic polynomial $p+7$, you can immediately write the general solution,



                          $$z=ce^{-7t}-2.$$



                          Now to get $y$, you integrate once, and obtain



                          $$y=c_0e^{-7t}-2t+c_1.$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 19 at 21:47









                          Yves Daoust

                          123k668218




                          123k668218






























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