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Grade: 12
        

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

9 years ago

Answers : (3)

Vijay Luxmi Askiitiansexpert
357 Points
							

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 -> .

9 years ago
Vishal Vaibhav
6 Points
							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.

						
9 years ago
Vishal Vaibhav
6 Points
							(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.

						
9 years ago
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