Showing that a second order differential equation has unique bounded solution











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I'm trying to show that $x''+bx'+x=cos(t)$ has a unique bounded solution, for $b<0$, $binmathbb{R}$. I believe I understand that it has a unique solution as all coefficients and the non-homogeneous term are all continuous, but the problem I'm having is the bounded part. I have that the three general solutions are:



$$x(t)=c_1e^{frac{-b+sqrt{b^2-4}}{2}t}+c_2e^{frac{-b-sqrt{b^2-4}}{2}t}+frac{sin(t)}{b},thinspacethinspace b^2-4>0 $$
$$x(t)=c_1e^t+c_2te^t-frac{sin(t)}{2},thinspacethinspace b^2-4=0 implies b=-2$$
$$x(t)=e^{frac{-b}{2}t}left(c_1cosleft(frac{sqrt{4-b^2}}{2}tright)+c_2sinleft(frac{sqrt{4-b^2}}{2}tright)right)+frac{sin(t)}{b},thinspacethinspace b^2-4<0$$



The only thing that I can see is that if for all cases, the solution is only bounded if $c_1=c_2=0$, seeing as the exponential term is always growing. Am I missing something about boundedness? Thank you for your time!










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    down vote

    favorite












    I'm trying to show that $x''+bx'+x=cos(t)$ has a unique bounded solution, for $b<0$, $binmathbb{R}$. I believe I understand that it has a unique solution as all coefficients and the non-homogeneous term are all continuous, but the problem I'm having is the bounded part. I have that the three general solutions are:



    $$x(t)=c_1e^{frac{-b+sqrt{b^2-4}}{2}t}+c_2e^{frac{-b-sqrt{b^2-4}}{2}t}+frac{sin(t)}{b},thinspacethinspace b^2-4>0 $$
    $$x(t)=c_1e^t+c_2te^t-frac{sin(t)}{2},thinspacethinspace b^2-4=0 implies b=-2$$
    $$x(t)=e^{frac{-b}{2}t}left(c_1cosleft(frac{sqrt{4-b^2}}{2}tright)+c_2sinleft(frac{sqrt{4-b^2}}{2}tright)right)+frac{sin(t)}{b},thinspacethinspace b^2-4<0$$



    The only thing that I can see is that if for all cases, the solution is only bounded if $c_1=c_2=0$, seeing as the exponential term is always growing. Am I missing something about boundedness? Thank you for your time!










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I'm trying to show that $x''+bx'+x=cos(t)$ has a unique bounded solution, for $b<0$, $binmathbb{R}$. I believe I understand that it has a unique solution as all coefficients and the non-homogeneous term are all continuous, but the problem I'm having is the bounded part. I have that the three general solutions are:



      $$x(t)=c_1e^{frac{-b+sqrt{b^2-4}}{2}t}+c_2e^{frac{-b-sqrt{b^2-4}}{2}t}+frac{sin(t)}{b},thinspacethinspace b^2-4>0 $$
      $$x(t)=c_1e^t+c_2te^t-frac{sin(t)}{2},thinspacethinspace b^2-4=0 implies b=-2$$
      $$x(t)=e^{frac{-b}{2}t}left(c_1cosleft(frac{sqrt{4-b^2}}{2}tright)+c_2sinleft(frac{sqrt{4-b^2}}{2}tright)right)+frac{sin(t)}{b},thinspacethinspace b^2-4<0$$



      The only thing that I can see is that if for all cases, the solution is only bounded if $c_1=c_2=0$, seeing as the exponential term is always growing. Am I missing something about boundedness? Thank you for your time!










      share|cite|improve this question















      I'm trying to show that $x''+bx'+x=cos(t)$ has a unique bounded solution, for $b<0$, $binmathbb{R}$. I believe I understand that it has a unique solution as all coefficients and the non-homogeneous term are all continuous, but the problem I'm having is the bounded part. I have that the three general solutions are:



      $$x(t)=c_1e^{frac{-b+sqrt{b^2-4}}{2}t}+c_2e^{frac{-b-sqrt{b^2-4}}{2}t}+frac{sin(t)}{b},thinspacethinspace b^2-4>0 $$
      $$x(t)=c_1e^t+c_2te^t-frac{sin(t)}{2},thinspacethinspace b^2-4=0 implies b=-2$$
      $$x(t)=e^{frac{-b}{2}t}left(c_1cosleft(frac{sqrt{4-b^2}}{2}tright)+c_2sinleft(frac{sqrt{4-b^2}}{2}tright)right)+frac{sin(t)}{b},thinspacethinspace b^2-4<0$$



      The only thing that I can see is that if for all cases, the solution is only bounded if $c_1=c_2=0$, seeing as the exponential term is always growing. Am I missing something about boundedness? Thank you for your time!







      differential-equations upper-lower-bounds






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      edited Nov 17 at 3:42









      Tianlalu

      2,814732




      2,814732










      asked Nov 17 at 3:35









      Leighmm

      32




      32






















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          You are correct.



          Since $b$ is the damping coefficient we get boundedness if $b>0$



          If $b>0$ then it is easy to see that the solutions are bounded.



          There is a typo in $b<0$ part.






          share|cite|improve this answer





















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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote



            accepted










            You are correct.



            Since $b$ is the damping coefficient we get boundedness if $b>0$



            If $b>0$ then it is easy to see that the solutions are bounded.



            There is a typo in $b<0$ part.






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              You are correct.



              Since $b$ is the damping coefficient we get boundedness if $b>0$



              If $b>0$ then it is easy to see that the solutions are bounded.



              There is a typo in $b<0$ part.






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                You are correct.



                Since $b$ is the damping coefficient we get boundedness if $b>0$



                If $b>0$ then it is easy to see that the solutions are bounded.



                There is a typo in $b<0$ part.






                share|cite|improve this answer












                You are correct.



                Since $b$ is the damping coefficient we get boundedness if $b>0$



                If $b>0$ then it is easy to see that the solutions are bounded.



                There is a typo in $b<0$ part.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 17 at 3:44









                Mohammad Riazi-Kermani

                40.3k41958




                40.3k41958






























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