Non bounded ODE solution












1












$begingroup$


Find the value of $w ge0$ so that the ODE below doesn't have a bounded solution



$$y''+y=cos(wt)$$



My attempt:



The solution should be in the form



$y(t)=Asin(wt)+Bcos(wt)$



$y'(t)=Awcos(wt)-Bwsin(wt)$



$y''(t)=-Aw^2sin(wt)-Bw^2cos(wt)$



So $(1): -Aw^2sin(wt)-Bw^2cos(wt)+Asin(wt)+Bcos(wt)=cos(wt)$



$Asin(wt)(1-w^2)+Bcos(wt)(1-w^2)=cos(wt)$ $Rightarrow A=0$ and $B=frac{1}{1-w^2}$



Thus $y(t)=frac{1}{1-w^2}cos(wt)$



Since $cosx$ is bounded for every $xin mathbb{R}$ does that mean that $w=1$ so that the solution is not bounded? Does not bounded means not defined in this case? Or am I very wrong?






ordinary-differential-equations initial-value-problems







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Find the value of $w ge0$ so that the ODE below doesn't have a bounded solution



    $$y''+y=cos(wt)$$



    My attempt:



    The solution should be in the form



    $y(t)=Asin(wt)+Bcos(wt)$



    $y'(t)=Awcos(wt)-Bwsin(wt)$



    $y''(t)=-Aw^2sin(wt)-Bw^2cos(wt)$



    So $(1): -Aw^2sin(wt)-Bw^2cos(wt)+Asin(wt)+Bcos(wt)=cos(wt)$



    $Asin(wt)(1-w^2)+Bcos(wt)(1-w^2)=cos(wt)$ $Rightarrow A=0$ and $B=frac{1}{1-w^2}$



    Thus $y(t)=frac{1}{1-w^2}cos(wt)$



    Since $cosx$ is bounded for every $xin mathbb{R}$ does that mean that $w=1$ so that the solution is not bounded? Does not bounded means not defined in this case? Or am I very wrong?






    ordinary-differential-equations initial-value-problems







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Find the value of $w ge0$ so that the ODE below doesn't have a bounded solution



      $$y''+y=cos(wt)$$



      My attempt:



      The solution should be in the form



      $y(t)=Asin(wt)+Bcos(wt)$



      $y'(t)=Awcos(wt)-Bwsin(wt)$



      $y''(t)=-Aw^2sin(wt)-Bw^2cos(wt)$



      So $(1): -Aw^2sin(wt)-Bw^2cos(wt)+Asin(wt)+Bcos(wt)=cos(wt)$



      $Asin(wt)(1-w^2)+Bcos(wt)(1-w^2)=cos(wt)$ $Rightarrow A=0$ and $B=frac{1}{1-w^2}$



      Thus $y(t)=frac{1}{1-w^2}cos(wt)$



      Since $cosx$ is bounded for every $xin mathbb{R}$ does that mean that $w=1$ so that the solution is not bounded? Does not bounded means not defined in this case? Or am I very wrong?






      ordinary-differential-equations initial-value-problems







      share|cite|improve this question









      $endgroup$




      Find the value of $w ge0$ so that the ODE below doesn't have a bounded solution



      $$y''+y=cos(wt)$$



      My attempt:



      The solution should be in the form



      $y(t)=Asin(wt)+Bcos(wt)$



      $y'(t)=Awcos(wt)-Bwsin(wt)$



      $y''(t)=-Aw^2sin(wt)-Bw^2cos(wt)$



      So $(1): -Aw^2sin(wt)-Bw^2cos(wt)+Asin(wt)+Bcos(wt)=cos(wt)$



      $Asin(wt)(1-w^2)+Bcos(wt)(1-w^2)=cos(wt)$ $Rightarrow A=0$ and $B=frac{1}{1-w^2}$



      Thus $y(t)=frac{1}{1-w^2}cos(wt)$



      Since $cosx$ is bounded for every $xin mathbb{R}$ does that mean that $w=1$ so that the solution is not bounded? Does not bounded means not defined in this case? Or am I very wrong?






      ordinary-differential-equations initial-value-problems




      ordinary-differential-equations initial-value-problems






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 26 at 0:49









      VakiPitsiVakiPitsi

      1978




      1978






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Your solution gives a particular solution, but not the whole solution. You have to add the solution to the corresponding homogeneous equation $y''+y=0$, which is easily seen to be $c_1 cos(t) + c_2 sin(t)$. Hence, $$y(t)=c_1 cos(t) + c_2 sin(t) + frac{1}{1-w^2}cos(wt)$$



          But this is still bounded for all $t$. However, notice that this solution is valid for $w neq 1$, but not for $w=1$ since you would obtain $0=cos(wt)$ when you plug derivatives back into the differential equation if $w=1$.



          For $w=1$, you may guess the solution to be of the form $A cos(t) + B sin(t)$, but it can't be since these are already included in the solution to the homogeneous equation. The strategy here is to assume the solution has the form $A t cos(t) + B t sin(t)$. Then,



          $$begin{cases} y=At sin(t)+Bt cos(t) \ y'=At cos(t)+B cos(t)-Bt sin(t) +A sin(t) \ y''= -Bt cos(t)+2A cos(t)-Atsin(t)-2B sin(t) end{cases}$$
          $$implies left( -Bt cos(t)+2A cos(t)-Atsin(t)-2B sin(t) right) + left( At sin(t)+Bt cos(t) right) = cos(t)$$
          $$implies 2A cos(t) - 2B sin(t) = cos(t) implies begin{cases} A=frac{1}{2} \ B =0 end{cases}$$



          So we obtain the full solution for when $w=1$ as



          $$y(t)=c_1 cos(t) + c_2 sin(t) + frac12tsin(t)$$



          which you can see is unbounded.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your comment! $y(t)=c_1cos(t)+c_2sin(t)+frac{1}{2}tsin(t)$ is unbounded because $t$ can take any value right?
            $endgroup$
            – VakiPitsi
            Jan 26 at 13:20












          • $begingroup$
            Yes, that’s right.
            $endgroup$
            – greelious
            Jan 26 at 14:09



















          0












          $begingroup$

          Compare the solution provided to that obtained by use of the Laplace transform. Using $y(t) to y(s)$ and $y''(t) to s^2 , y(s) - y'(0) - y(0) , s$ then
          begin{align}
          s^2 , y(s) - y'(0) - y(0) , s + y(s) &= frac{s}{s^2 + omega^2} \
          y(s) &= frac{y'(0)}{s^2 + 1} + frac{y(0) , s}{s^2 + 1} + frac{s}{(s^2 +1)(s^2 + omega^2)} \
          y(t) &= y(0) , cos(t) + y'(0) , sin(t) - frac{cos(t) - cos(omega t)}{1-omega^2} \
          &= left( y(0) -frac{1}{1-omega^2} right) , cos(t) + y'(0) , sin(t) + frac{cos(omega t)}{1- omega^2}.
          end{align}



          This solution is in agreement with that of the proposer and the Laplace transform used is
          $$y(s) = int_{0}^{infty} e^{- s t} , y(t) , dt.$$






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Your solution gives a particular solution, but not the whole solution. You have to add the solution to the corresponding homogeneous equation $y''+y=0$, which is easily seen to be $c_1 cos(t) + c_2 sin(t)$. Hence, $$y(t)=c_1 cos(t) + c_2 sin(t) + frac{1}{1-w^2}cos(wt)$$



            But this is still bounded for all $t$. However, notice that this solution is valid for $w neq 1$, but not for $w=1$ since you would obtain $0=cos(wt)$ when you plug derivatives back into the differential equation if $w=1$.



            For $w=1$, you may guess the solution to be of the form $A cos(t) + B sin(t)$, but it can't be since these are already included in the solution to the homogeneous equation. The strategy here is to assume the solution has the form $A t cos(t) + B t sin(t)$. Then,



            $$begin{cases} y=At sin(t)+Bt cos(t) \ y'=At cos(t)+B cos(t)-Bt sin(t) +A sin(t) \ y''= -Bt cos(t)+2A cos(t)-Atsin(t)-2B sin(t) end{cases}$$
            $$implies left( -Bt cos(t)+2A cos(t)-Atsin(t)-2B sin(t) right) + left( At sin(t)+Bt cos(t) right) = cos(t)$$
            $$implies 2A cos(t) - 2B sin(t) = cos(t) implies begin{cases} A=frac{1}{2} \ B =0 end{cases}$$



            So we obtain the full solution for when $w=1$ as



            $$y(t)=c_1 cos(t) + c_2 sin(t) + frac12tsin(t)$$



            which you can see is unbounded.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you for your comment! $y(t)=c_1cos(t)+c_2sin(t)+frac{1}{2}tsin(t)$ is unbounded because $t$ can take any value right?
              $endgroup$
              – VakiPitsi
              Jan 26 at 13:20












            • $begingroup$
              Yes, that’s right.
              $endgroup$
              – greelious
              Jan 26 at 14:09
















            1












            $begingroup$

            Your solution gives a particular solution, but not the whole solution. You have to add the solution to the corresponding homogeneous equation $y''+y=0$, which is easily seen to be $c_1 cos(t) + c_2 sin(t)$. Hence, $$y(t)=c_1 cos(t) + c_2 sin(t) + frac{1}{1-w^2}cos(wt)$$



            But this is still bounded for all $t$. However, notice that this solution is valid for $w neq 1$, but not for $w=1$ since you would obtain $0=cos(wt)$ when you plug derivatives back into the differential equation if $w=1$.



            For $w=1$, you may guess the solution to be of the form $A cos(t) + B sin(t)$, but it can't be since these are already included in the solution to the homogeneous equation. The strategy here is to assume the solution has the form $A t cos(t) + B t sin(t)$. Then,



            $$begin{cases} y=At sin(t)+Bt cos(t) \ y'=At cos(t)+B cos(t)-Bt sin(t) +A sin(t) \ y''= -Bt cos(t)+2A cos(t)-Atsin(t)-2B sin(t) end{cases}$$
            $$implies left( -Bt cos(t)+2A cos(t)-Atsin(t)-2B sin(t) right) + left( At sin(t)+Bt cos(t) right) = cos(t)$$
            $$implies 2A cos(t) - 2B sin(t) = cos(t) implies begin{cases} A=frac{1}{2} \ B =0 end{cases}$$



            So we obtain the full solution for when $w=1$ as



            $$y(t)=c_1 cos(t) + c_2 sin(t) + frac12tsin(t)$$



            which you can see is unbounded.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you for your comment! $y(t)=c_1cos(t)+c_2sin(t)+frac{1}{2}tsin(t)$ is unbounded because $t$ can take any value right?
              $endgroup$
              – VakiPitsi
              Jan 26 at 13:20












            • $begingroup$
              Yes, that’s right.
              $endgroup$
              – greelious
              Jan 26 at 14:09














            1












            1








            1





            $begingroup$

            Your solution gives a particular solution, but not the whole solution. You have to add the solution to the corresponding homogeneous equation $y''+y=0$, which is easily seen to be $c_1 cos(t) + c_2 sin(t)$. Hence, $$y(t)=c_1 cos(t) + c_2 sin(t) + frac{1}{1-w^2}cos(wt)$$



            But this is still bounded for all $t$. However, notice that this solution is valid for $w neq 1$, but not for $w=1$ since you would obtain $0=cos(wt)$ when you plug derivatives back into the differential equation if $w=1$.



            For $w=1$, you may guess the solution to be of the form $A cos(t) + B sin(t)$, but it can't be since these are already included in the solution to the homogeneous equation. The strategy here is to assume the solution has the form $A t cos(t) + B t sin(t)$. Then,



            $$begin{cases} y=At sin(t)+Bt cos(t) \ y'=At cos(t)+B cos(t)-Bt sin(t) +A sin(t) \ y''= -Bt cos(t)+2A cos(t)-Atsin(t)-2B sin(t) end{cases}$$
            $$implies left( -Bt cos(t)+2A cos(t)-Atsin(t)-2B sin(t) right) + left( At sin(t)+Bt cos(t) right) = cos(t)$$
            $$implies 2A cos(t) - 2B sin(t) = cos(t) implies begin{cases} A=frac{1}{2} \ B =0 end{cases}$$



            So we obtain the full solution for when $w=1$ as



            $$y(t)=c_1 cos(t) + c_2 sin(t) + frac12tsin(t)$$



            which you can see is unbounded.






            share|cite|improve this answer











            $endgroup$



            Your solution gives a particular solution, but not the whole solution. You have to add the solution to the corresponding homogeneous equation $y''+y=0$, which is easily seen to be $c_1 cos(t) + c_2 sin(t)$. Hence, $$y(t)=c_1 cos(t) + c_2 sin(t) + frac{1}{1-w^2}cos(wt)$$



            But this is still bounded for all $t$. However, notice that this solution is valid for $w neq 1$, but not for $w=1$ since you would obtain $0=cos(wt)$ when you plug derivatives back into the differential equation if $w=1$.



            For $w=1$, you may guess the solution to be of the form $A cos(t) + B sin(t)$, but it can't be since these are already included in the solution to the homogeneous equation. The strategy here is to assume the solution has the form $A t cos(t) + B t sin(t)$. Then,



            $$begin{cases} y=At sin(t)+Bt cos(t) \ y'=At cos(t)+B cos(t)-Bt sin(t) +A sin(t) \ y''= -Bt cos(t)+2A cos(t)-Atsin(t)-2B sin(t) end{cases}$$
            $$implies left( -Bt cos(t)+2A cos(t)-Atsin(t)-2B sin(t) right) + left( At sin(t)+Bt cos(t) right) = cos(t)$$
            $$implies 2A cos(t) - 2B sin(t) = cos(t) implies begin{cases} A=frac{1}{2} \ B =0 end{cases}$$



            So we obtain the full solution for when $w=1$ as



            $$y(t)=c_1 cos(t) + c_2 sin(t) + frac12tsin(t)$$



            which you can see is unbounded.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 26 at 2:12

























            answered Jan 26 at 2:06









            greeliousgreelious

            472112




            472112












            • $begingroup$
              Thank you for your comment! $y(t)=c_1cos(t)+c_2sin(t)+frac{1}{2}tsin(t)$ is unbounded because $t$ can take any value right?
              $endgroup$
              – VakiPitsi
              Jan 26 at 13:20












            • $begingroup$
              Yes, that’s right.
              $endgroup$
              – greelious
              Jan 26 at 14:09


















            • $begingroup$
              Thank you for your comment! $y(t)=c_1cos(t)+c_2sin(t)+frac{1}{2}tsin(t)$ is unbounded because $t$ can take any value right?
              $endgroup$
              – VakiPitsi
              Jan 26 at 13:20












            • $begingroup$
              Yes, that’s right.
              $endgroup$
              – greelious
              Jan 26 at 14:09
















            $begingroup$
            Thank you for your comment! $y(t)=c_1cos(t)+c_2sin(t)+frac{1}{2}tsin(t)$ is unbounded because $t$ can take any value right?
            $endgroup$
            – VakiPitsi
            Jan 26 at 13:20






            $begingroup$
            Thank you for your comment! $y(t)=c_1cos(t)+c_2sin(t)+frac{1}{2}tsin(t)$ is unbounded because $t$ can take any value right?
            $endgroup$
            – VakiPitsi
            Jan 26 at 13:20














            $begingroup$
            Yes, that’s right.
            $endgroup$
            – greelious
            Jan 26 at 14:09




            $begingroup$
            Yes, that’s right.
            $endgroup$
            – greelious
            Jan 26 at 14:09











            0












            $begingroup$

            Compare the solution provided to that obtained by use of the Laplace transform. Using $y(t) to y(s)$ and $y''(t) to s^2 , y(s) - y'(0) - y(0) , s$ then
            begin{align}
            s^2 , y(s) - y'(0) - y(0) , s + y(s) &= frac{s}{s^2 + omega^2} \
            y(s) &= frac{y'(0)}{s^2 + 1} + frac{y(0) , s}{s^2 + 1} + frac{s}{(s^2 +1)(s^2 + omega^2)} \
            y(t) &= y(0) , cos(t) + y'(0) , sin(t) - frac{cos(t) - cos(omega t)}{1-omega^2} \
            &= left( y(0) -frac{1}{1-omega^2} right) , cos(t) + y'(0) , sin(t) + frac{cos(omega t)}{1- omega^2}.
            end{align}



            This solution is in agreement with that of the proposer and the Laplace transform used is
            $$y(s) = int_{0}^{infty} e^{- s t} , y(t) , dt.$$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Compare the solution provided to that obtained by use of the Laplace transform. Using $y(t) to y(s)$ and $y''(t) to s^2 , y(s) - y'(0) - y(0) , s$ then
              begin{align}
              s^2 , y(s) - y'(0) - y(0) , s + y(s) &= frac{s}{s^2 + omega^2} \
              y(s) &= frac{y'(0)}{s^2 + 1} + frac{y(0) , s}{s^2 + 1} + frac{s}{(s^2 +1)(s^2 + omega^2)} \
              y(t) &= y(0) , cos(t) + y'(0) , sin(t) - frac{cos(t) - cos(omega t)}{1-omega^2} \
              &= left( y(0) -frac{1}{1-omega^2} right) , cos(t) + y'(0) , sin(t) + frac{cos(omega t)}{1- omega^2}.
              end{align}



              This solution is in agreement with that of the proposer and the Laplace transform used is
              $$y(s) = int_{0}^{infty} e^{- s t} , y(t) , dt.$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Compare the solution provided to that obtained by use of the Laplace transform. Using $y(t) to y(s)$ and $y''(t) to s^2 , y(s) - y'(0) - y(0) , s$ then
                begin{align}
                s^2 , y(s) - y'(0) - y(0) , s + y(s) &= frac{s}{s^2 + omega^2} \
                y(s) &= frac{y'(0)}{s^2 + 1} + frac{y(0) , s}{s^2 + 1} + frac{s}{(s^2 +1)(s^2 + omega^2)} \
                y(t) &= y(0) , cos(t) + y'(0) , sin(t) - frac{cos(t) - cos(omega t)}{1-omega^2} \
                &= left( y(0) -frac{1}{1-omega^2} right) , cos(t) + y'(0) , sin(t) + frac{cos(omega t)}{1- omega^2}.
                end{align}



                This solution is in agreement with that of the proposer and the Laplace transform used is
                $$y(s) = int_{0}^{infty} e^{- s t} , y(t) , dt.$$






                share|cite|improve this answer









                $endgroup$



                Compare the solution provided to that obtained by use of the Laplace transform. Using $y(t) to y(s)$ and $y''(t) to s^2 , y(s) - y'(0) - y(0) , s$ then
                begin{align}
                s^2 , y(s) - y'(0) - y(0) , s + y(s) &= frac{s}{s^2 + omega^2} \
                y(s) &= frac{y'(0)}{s^2 + 1} + frac{y(0) , s}{s^2 + 1} + frac{s}{(s^2 +1)(s^2 + omega^2)} \
                y(t) &= y(0) , cos(t) + y'(0) , sin(t) - frac{cos(t) - cos(omega t)}{1-omega^2} \
                &= left( y(0) -frac{1}{1-omega^2} right) , cos(t) + y'(0) , sin(t) + frac{cos(omega t)}{1- omega^2}.
                end{align}



                This solution is in agreement with that of the proposer and the Laplace transform used is
                $$y(s) = int_{0}^{infty} e^{- s t} , y(t) , dt.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 26 at 2:28









                LeucippusLeucippus

                19.7k102871




                19.7k102871






























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                    How to fix TextFormField cause rebuild widget in Flutter

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                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
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                    Npm cannot find a required file even through it is in the searched directory

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                    up vote 0 down vote favorite I got that error: users-Air:pdfuploader user$ npm run start client@0.1.0 start /Users/user/Documents/pdfuploader react-scripts start Could not find a required file. Name: index.html Searched in: /Users/user/Documents/pdfuploader/public npm ERR! code ELIFECYCLE npm ERR! errno 1 npm ERR! client@0.1.0 start: `react-scripts start` npm ERR! Exit status 1 npm ERR! npm ERR! Failed at the client@0.1.0 start script. npm ERR! This is probably not a problem with npm. There is likely additional logging output above. npm ERR! A complete log of this run can be found in: npm ERR! /Users/user/.npm/_logs/2018-11-19T10_19_14_555Z-debug.log users-Air:pdfuploader user$ And the file is in /Users/user/Documents/pdfuploader/public as you can see in my github repository: https://github.c...
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