let y=y(t) be a solution to the differential equation y'+2ty=t^2,then lim(t tends to infinity) [y/t] is?

Given differential equation is,

                        dy/dt + 2t y = t²

         This is a linear differential equation where P(t) = 2t and Q(t) = t²,

        So, the solution of this differential equation will be given by

        After solving this, you need to simply find greatest integer of y/t, i.e.,[y/t]  and take limit t -> .

Answers is y/t=1/2+O(1/t) so that [y/t]=0. (O is the order symbol) Heuristic method: putting y=u*t we have u'+(2t^2+1)u=t^2. assuming u and u' are finite as t-> infinity we have on dividing by t^2 0+(2+0)u_infty=1. hence [u]_infty=0. Rigorous Proof: (determination of the asymptotic solution) multiplying by exp(t^2) both sides of the equation one gets the following solution (using LATEX notation) $$y=C\exp (-t^2)+\exp (-t^2){\int }_0^t{\tau }^{2}exp(-{\tau }^2)d\tau .$$ where C is the constant of integration. On integrating by parts one gets $$y=C\exp (-t^2)+t/2-(1/2)\exp (-t^2){\int }_0^{t}exp(-{\tau }^2)d\tau .$$ Use the substitution $$\tau =t(1-\xi )$$ to obtain $$y=C\exp (-t^2)+t/2-(t/2){\int }_0^{1}exp(-2t\xi +t{\xi }^2)d\xi .$$ Let us consider the integral $$I={\int }_0^{1}exp(-2t\xi +t{\xi }^2)d\xi <{\int }_0^{1}exp(-t\xi )d\xi =(1-\exp (-t))/t$$ as \xi^2<\xi for \xi \in [0,1]. From this inequality and on dividing by t both sides we have $$y/t=1/2+O(1/t).$$ so that [y/t]=0. Q.E.D.

(Corrected some typos in the earlier post!) Answers is y/t=1/2+O(1/t) so that [y/t]=0 as t--> infty. (O is the order symbol) Heuristic method: putting y=u*t we have u'+(2t^2+1)u=t^2. assuming u and u' are finite as t-> infinity we have on dividing by t^2 0+(2+0)u_infty=1. hence [u]_infty=0. Rigorous Proof: (determination of the asymptotic solution) multiplying by exp(t^2) both sides of the equation one gets the following solution (using LATEX notation) $$y=C\exp (-t^2)+\exp (-t^2){\int }_0^t{\tau }^{2}exp({\tau }^2)d\tau .$$ where C is the constant of integration. On integrating by parts one gets $$y=C\exp (-t^2)+t/2-(1/2)\exp (-t^2){\int }_0^{t}exp({\tau }^2)d\tau .$$ Use the substitution $$\tau =t(1-\xi )$$ to obtain $$y=C\exp (-t^2)+t/2-(t/2){\int }_0^{1}exp(-2t\xi +t{\xi }^2)d\xi .$$ Let us consider the integral $$I={\int }_0^{1}exp(-2t\xi +t{\xi }^2)d\xi <{\int }_0^{1}exp(-t\xi )d\xi =(1-\exp (-t))/t$$ as \xi^2<\xi for \xi \in [0,1]. From this inequality and on dividing by t both sides we have $$y/t=1/2+O(1/t).$$ so that [y/t]=0 as t--> infty. Q.E.D.