write a differential equation $(frac{dy}{dt} = ay+b)$ whose solutions have the required behavior as $...












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Let all other solutions diverge from $y = frac{3}{4}$. write a differential equation $(frac{dy}{dt} = ay +b)$ whose solutions have the required behavior as t goes to infinity. Explain how you came up with the differential equation).



I think the answer is $y' = 4y - 3$ but I am not sure. Can someone help me produce the answer with complete working?










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    1












    $begingroup$


    Let all other solutions diverge from $y = frac{3}{4}$. write a differential equation $(frac{dy}{dt} = ay +b)$ whose solutions have the required behavior as t goes to infinity. Explain how you came up with the differential equation).



    I think the answer is $y' = 4y - 3$ but I am not sure. Can someone help me produce the answer with complete working?










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let all other solutions diverge from $y = frac{3}{4}$. write a differential equation $(frac{dy}{dt} = ay +b)$ whose solutions have the required behavior as t goes to infinity. Explain how you came up with the differential equation).



      I think the answer is $y' = 4y - 3$ but I am not sure. Can someone help me produce the answer with complete working?










      share|cite|improve this question











      $endgroup$




      Let all other solutions diverge from $y = frac{3}{4}$. write a differential equation $(frac{dy}{dt} = ay +b)$ whose solutions have the required behavior as t goes to infinity. Explain how you came up with the differential equation).



      I think the answer is $y' = 4y - 3$ but I am not sure. Can someone help me produce the answer with complete working?







      ordinary-differential-equations






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 21 at 5:59









      user549397

      1,5061418




      1,5061418










      asked Sep 22 '16 at 12:07









      AnonymousAnonymous

      214




      214






















          2 Answers
          2






          active

          oldest

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          $begingroup$

          well i think you are really close to the solution:



          $$y'=-4y-3$$



          Homogeneous part would be:
          $$y'+ 4y = 0$$
          which leads to:
          $$y0=Ce^{-4t}$$



          particular part would be (with initial assumption $yp=ax+b$):
          $$yp=3/4$$



          which leads to:
          $$y(t)=Ce^{-4t}+3/4$$



          $$limlimits_{t to infty} y(t)=3/4$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is it y' = -4y - 3 or y' = 4y - 3
            $endgroup$
            – Anonymous
            Sep 22 '16 at 13:03












          • $begingroup$
            y'= -4y-3 try to solve some similar equations.
            $endgroup$
            – arash javan
            Sep 22 '16 at 13:07












          • $begingroup$
            @Anonymous mark the answer plz, if it was your solution
            $endgroup$
            – arash javan
            Sep 23 '16 at 14:09



















          0












          $begingroup$

          Differential equation $y' = ax +b = f(y)$ is a nonautonuous equation. If a = 0, this equation becomes $y' = b$. Otherwise, if $a ne 0$, the equation has a unique equibrium point $y^* = -b/a$ and $f'(y^*) = a$. Therefore, by Lyapunov indirect method, we conclude that



          If $a > 0$ then $y^*$ is unstable.



          If $a < 0$ then $y^*$ is globally asymptotically stable. It implies that all solutions converge to $y^*$ as $t to infty$.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            well i think you are really close to the solution:



            $$y'=-4y-3$$



            Homogeneous part would be:
            $$y'+ 4y = 0$$
            which leads to:
            $$y0=Ce^{-4t}$$



            particular part would be (with initial assumption $yp=ax+b$):
            $$yp=3/4$$



            which leads to:
            $$y(t)=Ce^{-4t}+3/4$$



            $$limlimits_{t to infty} y(t)=3/4$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Is it y' = -4y - 3 or y' = 4y - 3
              $endgroup$
              – Anonymous
              Sep 22 '16 at 13:03












            • $begingroup$
              y'= -4y-3 try to solve some similar equations.
              $endgroup$
              – arash javan
              Sep 22 '16 at 13:07












            • $begingroup$
              @Anonymous mark the answer plz, if it was your solution
              $endgroup$
              – arash javan
              Sep 23 '16 at 14:09
















            0












            $begingroup$

            well i think you are really close to the solution:



            $$y'=-4y-3$$



            Homogeneous part would be:
            $$y'+ 4y = 0$$
            which leads to:
            $$y0=Ce^{-4t}$$



            particular part would be (with initial assumption $yp=ax+b$):
            $$yp=3/4$$



            which leads to:
            $$y(t)=Ce^{-4t}+3/4$$



            $$limlimits_{t to infty} y(t)=3/4$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Is it y' = -4y - 3 or y' = 4y - 3
              $endgroup$
              – Anonymous
              Sep 22 '16 at 13:03












            • $begingroup$
              y'= -4y-3 try to solve some similar equations.
              $endgroup$
              – arash javan
              Sep 22 '16 at 13:07












            • $begingroup$
              @Anonymous mark the answer plz, if it was your solution
              $endgroup$
              – arash javan
              Sep 23 '16 at 14:09














            0












            0








            0





            $begingroup$

            well i think you are really close to the solution:



            $$y'=-4y-3$$



            Homogeneous part would be:
            $$y'+ 4y = 0$$
            which leads to:
            $$y0=Ce^{-4t}$$



            particular part would be (with initial assumption $yp=ax+b$):
            $$yp=3/4$$



            which leads to:
            $$y(t)=Ce^{-4t}+3/4$$



            $$limlimits_{t to infty} y(t)=3/4$$






            share|cite|improve this answer











            $endgroup$



            well i think you are really close to the solution:



            $$y'=-4y-3$$



            Homogeneous part would be:
            $$y'+ 4y = 0$$
            which leads to:
            $$y0=Ce^{-4t}$$



            particular part would be (with initial assumption $yp=ax+b$):
            $$yp=3/4$$



            which leads to:
            $$y(t)=Ce^{-4t}+3/4$$



            $$limlimits_{t to infty} y(t)=3/4$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Sep 22 '16 at 13:00

























            answered Sep 22 '16 at 12:46









            arash javanarash javan

            1397




            1397












            • $begingroup$
              Is it y' = -4y - 3 or y' = 4y - 3
              $endgroup$
              – Anonymous
              Sep 22 '16 at 13:03












            • $begingroup$
              y'= -4y-3 try to solve some similar equations.
              $endgroup$
              – arash javan
              Sep 22 '16 at 13:07












            • $begingroup$
              @Anonymous mark the answer plz, if it was your solution
              $endgroup$
              – arash javan
              Sep 23 '16 at 14:09


















            • $begingroup$
              Is it y' = -4y - 3 or y' = 4y - 3
              $endgroup$
              – Anonymous
              Sep 22 '16 at 13:03












            • $begingroup$
              y'= -4y-3 try to solve some similar equations.
              $endgroup$
              – arash javan
              Sep 22 '16 at 13:07












            • $begingroup$
              @Anonymous mark the answer plz, if it was your solution
              $endgroup$
              – arash javan
              Sep 23 '16 at 14:09
















            $begingroup$
            Is it y' = -4y - 3 or y' = 4y - 3
            $endgroup$
            – Anonymous
            Sep 22 '16 at 13:03






            $begingroup$
            Is it y' = -4y - 3 or y' = 4y - 3
            $endgroup$
            – Anonymous
            Sep 22 '16 at 13:03














            $begingroup$
            y'= -4y-3 try to solve some similar equations.
            $endgroup$
            – arash javan
            Sep 22 '16 at 13:07






            $begingroup$
            y'= -4y-3 try to solve some similar equations.
            $endgroup$
            – arash javan
            Sep 22 '16 at 13:07














            $begingroup$
            @Anonymous mark the answer plz, if it was your solution
            $endgroup$
            – arash javan
            Sep 23 '16 at 14:09




            $begingroup$
            @Anonymous mark the answer plz, if it was your solution
            $endgroup$
            – arash javan
            Sep 23 '16 at 14:09











            0












            $begingroup$

            Differential equation $y' = ax +b = f(y)$ is a nonautonuous equation. If a = 0, this equation becomes $y' = b$. Otherwise, if $a ne 0$, the equation has a unique equibrium point $y^* = -b/a$ and $f'(y^*) = a$. Therefore, by Lyapunov indirect method, we conclude that



            If $a > 0$ then $y^*$ is unstable.



            If $a < 0$ then $y^*$ is globally asymptotically stable. It implies that all solutions converge to $y^*$ as $t to infty$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Differential equation $y' = ax +b = f(y)$ is a nonautonuous equation. If a = 0, this equation becomes $y' = b$. Otherwise, if $a ne 0$, the equation has a unique equibrium point $y^* = -b/a$ and $f'(y^*) = a$. Therefore, by Lyapunov indirect method, we conclude that



              If $a > 0$ then $y^*$ is unstable.



              If $a < 0$ then $y^*$ is globally asymptotically stable. It implies that all solutions converge to $y^*$ as $t to infty$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Differential equation $y' = ax +b = f(y)$ is a nonautonuous equation. If a = 0, this equation becomes $y' = b$. Otherwise, if $a ne 0$, the equation has a unique equibrium point $y^* = -b/a$ and $f'(y^*) = a$. Therefore, by Lyapunov indirect method, we conclude that



                If $a > 0$ then $y^*$ is unstable.



                If $a < 0$ then $y^*$ is globally asymptotically stable. It implies that all solutions converge to $y^*$ as $t to infty$.






                share|cite|improve this answer









                $endgroup$



                Differential equation $y' = ax +b = f(y)$ is a nonautonuous equation. If a = 0, this equation becomes $y' = b$. Otherwise, if $a ne 0$, the equation has a unique equibrium point $y^* = -b/a$ and $f'(y^*) = a$. Therefore, by Lyapunov indirect method, we conclude that



                If $a > 0$ then $y^*$ is unstable.



                If $a < 0$ then $y^*$ is globally asymptotically stable. It implies that all solutions converge to $y^*$ as $t to infty$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 17 '18 at 19:12









                MichaelCarrickMichaelCarrick

                1097




                1097






























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