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

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

Grade:12

3 Answers

Vijay Luxmi Askiitiansexpert
357 Points
12 years ago

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 -> .

Vishal Vaibhav
6 Points
12 years ago
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.
Vishal Vaibhav
6 Points
12 years ago
(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.

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