Click to Chat

0120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list

Get extra R 400 off
USE CODE: SSPD25

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


7 years ago

Share



<br/><!--
<br/> /* Font Definitions */
<br/> @font-face
<br/>	{font-family:"Cambria Math";
<br/>	panose-1:2 4 5 3 5 4 6 3 2 4;
<br/>	mso-font-charset:0;
<br/>	mso-generic-font-family:roman;
<br/>	mso-font-pitch:variable;
<br/>	mso-font-signature:-1610611985 1107304683 0 0 159 0;}
<br/>@font-face
<br/>	{font-family:Calibri;
<br/>	panose-1:2 15 5 2 2 2 4 3 2 4;
<br/>	mso-font-charset:0;
<br/>	mso-generic-font-family:swiss;
<br/>	mso-font-pitch:variable;
<br/>	mso-font-signature:-1610611985 1073750139 0 0 159 0;}
<br/>@font-face
<br/>	{font-family:Verdana;
<br/>	panose-1:2 11 6 4 3 5 4 4 2 4;
<br/>	mso-font-charset:0;
<br/>	mso-generic-font-family:swiss;
<br/>	mso-font-pitch:variable;
<br/>	mso-font-signature:536871559 0 0 0 415 0;}
<br/> /* Style Definitions */
<br/> p.MsoNormal, li.MsoNormal, div.MsoNormal
<br/>	{mso-style-unhide:no;
<br/>	mso-style-qformat:yes;
<br/>	mso-style-parent:"";
<br/>	margin-top:0in;
<br/>	margin-right:0in;
<br/>	margin-bottom:10.0pt;
<br/>	margin-left:0in;
<br/>	line-height:115%;
<br/>	mso-pagination:widow-orphan;
<br/>	font-size:11.0pt;
<br/>	font-family:"Calibri","sans-serif";
<br/>	mso-ascii-font-family:Calibri;
<br/>	mso-ascii-theme-font:minor-latin;
<br/>	mso-fareast-font-family:Calibri;
<br/>	mso-fareast-theme-font:minor-latin;
<br/>	mso-hansi-font-family:Calibri;
<br/>	mso-hansi-theme-font:minor-latin;
<br/>	mso-bidi-font-family:"Times New Roman";
<br/>	mso-bidi-theme-font:minor-bidi;}
<br/>.MsoChpDefault
<br/>	{mso-style-type:export-only;
<br/>	mso-default-props:yes;
<br/>	mso-ascii-font-family:Calibri;
<br/>	mso-ascii-theme-font:minor-latin;
<br/>	mso-fareast-font-family:Calibri;
<br/>	mso-fareast-theme-font:minor-latin;
<br/>	mso-hansi-font-family:Calibri;
<br/>	mso-hansi-theme-font:minor-latin;
<br/>	mso-bidi-font-family:"Times New Roman";
<br/>	mso-bidi-theme-font:minor-bidi;}
<br/>.MsoPapDefault
<br/>	{mso-style-type:export-only;
<br/>	margin-bottom:10.0pt;
<br/>	line-height:115%;}
<br/>@page Section1
<br/>	{size:8.5in 11.0in;
<br/>	margin:1.0in 1.0in 1.0in 1.0in;
<br/>	mso-footer-margin:.5in;
<br/>	mso-paper-source:0;}
<br/>div.Section1
<br/>	{page:Section1;}
<br/>-->
<br/>
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 -> .


7 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.


7 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.


7 years ago

# Other Related Questions on Integral Calculus

The area of loop of the curve ay^2=x^2(a-x) is 15a^2/p,then p=?

Hii try to divide it along any line passing through centre and then do the double integration of figure with respect to x and y . You will get answer in this case by this method

 Sourabh Singh 2 months ago
How many distinct values of 𝑥 satisfy the equation sin(𝑥) = 𝑥/2, where 𝑥 is in radians? (A) 1 (B) 2 (C) 3 (D) 4 or more

plot the graph of sin(x) and x/2. From just observing it we can see that it intersects at 3 points and hence the answer is 3.

 Riddhish Bhalodia 7 months ago

they give sin(x)=x/2 0 is the one solution because sin(0)=0 so one solution exist for this question if you take pi/2 or pi/4 you have to p[rove that sin pi/2=pi/4 or sin pi/8=pi/16 which is ...

 SREEKANTH 4 months ago
The number of 6 - digit number s that can be made with the digits 0,1,2,3,4&5 so that even digits occupy odd places is ( A ) 24 (B) 36 ( C ) 48 ( D ) NONE

the numbers which are allowed are 0,1,2,3,4,5 and the number has to be 6 digit ( _ _ _ _ _ _ ) so 0 cannot occupy lacks place (the number should not start with 0 as 0_ _ _ _ _ ) as it will...

 Ankit Jaiswal 2 months ago

i suppose i dont need to metion that 3!=3*2*1=6

 Ankit Jaiswal 2 months ago

 T Jahnavi 2 months ago
The sum of first 9 terms of the series 1³/1 +1³+2³/1+3...... Is

[hint: find the nth term and then find its summation] Tn for the series is : Tn= {(n+1)^2 }/4 Then use the summation of this Tn with n from 1 to 9. ∑ Tn = 96. Hope this helps. Comment if...

 Anuj Shrivastav one month ago
Sir, this is my doubt please clarify, Let f:R->R be a function given by f(x+y) = f(x)+f(y) for all x,y belongs to R such that F91)=a, f(x)= Sir, Kindly give me the solution

Lim h->0 [ f(x+h) - f(x) ] / h = f(x) Lim h->0 [ f(x) + f(h) - f(x) ] / h = f(x) As f(x+y) = f(x) + f(y) Lim h->0 f(h) / h = f`(x) Transforming into differential eq. y/x = dy/dx (variable...

 2017 years ago
View all Questions »

• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: R 15,000
• View Details
Get extra R 3,750 off
USE CODE: SSPD25

Get extra R 400 off
USE CODE: SSPD25

More Questions On Integral Calculus