let y=y(t) be a solution to the differential equation y'+2ty=t^2,then lim(t tends to infinity) [y/t] is?
Neha Anand , 16 Years ago
Grade 12
3 Answers
Vijay Luxmi Askiitiansexpert
Last Activity: 16 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
Last Activity: 15 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)
where C is the constant of integration. On integrating by parts one gets
Use the substitution
to obtain
Let us consider the integral
as \xi^2<\xi for \xi \in [0,1].
From this inequality and on dividing by t both sides we have
so that [y/t]=0. Q.E.D.
Vishal Vaibhav
Last Activity: 15 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)
where C is the constant of integration. On integrating by parts one gets
Use the substitution
to obtain
Let us consider the integral
as \xi^2<\xi for \xi \in [0,1].
From this inequality and on dividing by t both sides we have
so that [y/t]=0 as t--> infty. Q.E.D.
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