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Higher Order Differential Equations-Practical Differential Equations-Assignment, Exercises of Applied Differential Equations

Dr. Bhim Rao Ambedkar UniversityApplied Differential Equations

This assignment is for Applied Differential Equations course. It was given by Albert Pinto at B R Ambedkar National Institute of Technology. It includes: Suspended, Spring, Stretched, Initial, Velocity, Equilibrium, Position, Displacement, Expression, Damping, Force

Typology: Exercises

2011/2012

Uploaded on 07/11/2012

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Applied Differential Equations (MT2043) Assignment 04
Submission Date: 18/05/2012
Instructor: Dr. Rashid Ali
Q1. A steel ball weighing 128 lb is suspended from a spring, whereupon the spring is stretched 2
ft from its natural length. The ball is started in motion with no initial velocity by displacing it 6 inch
above the equilibrium poition. Assuming no air resistance, find (a) an expression for the
displacement of the ball at any time t, and (b) the position of the ball at t = π / 2.
Ans. ft
4
1
)
12
( );4cos(
2
1
)( −=−=
π
ytty
Q2. A 10 kg mass is attached to a spring, stretching it 0.7 m from its natural length. The mass is
started in motion from the equilibrium position, with an initial velocity of 1 m/sec. in the upward
direction. Find the displacement if the force due to air resistance (damping force) is – 90y′.
Ans. 0lim ),(
5
1
)( 277
2
2
1=−=+= ∞→
−−−− yeeececty t
tttt
Q3. Refer to Q2, if the applied external force F(t) = 5sint, find the displacement at any time t.
Ans. )cos9sin139990(
500
172 tteey tt −++−= −−
Q4. Recall that, the Newton’s Law of Cooling states that the temperature T of an object changes
at a rate proportional to the difference between the temperatures of the object (T) and the
surrounding medium (Ts). That is, 0
)0()( TTTTk
dt
dT
s=−= . Using this law, solve the following
problems:
A. An object takes 40 minutes to cool from 30oC to 24oC in a room having the room-
temperature of 20oC. Find the temperature of the object at any time t. What will be the
temperature of the object after 15 minutes? How long will it take the object to cool down to
21oC?
Hint: the IVP is 24)40( ,30)0(),20( ==−= TTTk
dt
dT ,
Ans.
minutes 5.100 time the,21 when and C1.27)15(,1020 0
40
)4.0ln(
===+= ⎟
⎠
⎞
⎜
⎝
⎛
tTTeT
t
B. One liter of ice-cream at a temperature of –15oC is removed from the deep freezer and
placed in a room where the temperature is 20oC. If after 20 minutes the temperature of the
ice-cream is –10oC, how long will it take the ice-cream to reach a temperature of 0oC?
Ans. minutes 63.72 time the,0 when and ,3520 20
7ln6ln
==−= ⎟
⎠
⎞
⎜
⎝
⎛−
tTeT
t
C. A thermometer is taken outside from an air-conditioned room where the temperature is
21oC. After 1 minute it reads 27oC and after 2 min. it reads 30oC. (That is, T(1) = 27 and
T(2) = 30). Find the outside temperature Ts.
Ans. Ts = 33oC
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Applied Differential Equations (MT2043) Assignment 04

Submission Date: 18/05/

Q1. A steel ball weighing 128 lb is suspended from a spring, whereupon the spring is stretched 2 ft from its natural length. The ball is started in motion with no initial velocity by displacing it 6 inch above the equilibrium poition. Assuming no air resistance, find ( a ) an expression for the displacement of the ball at any time t , and ( b ) the position of the ball at t = π / 2. Ans. ft 4

cos( 4 ); ( 2

yt t y

Q2. A 10 kg mass is attached to a spring, stretching it 0.7 m from its natural length. The mass is started in motion from the equilibrium position, with an initial velocity of 1 m/sec. in the upward direction. Find the displacement if the force due to air resistance (damping force) is – 90 y′.

Ans. ( ), lim 0 5

y ( t )= c 1 e −^2 t + c 2 e −^7 t = e −^7 t − e −^2 t t →∞ y =

Q3. Refer to Q2 , if the applied external force F ( t ) = 5sin t , find the displacement at any time t.

Ans. ( 90 99 13 sin 9 cos) 500

y = − e − t^ + e − t + t − t

Q4. Recall that, the Newton’s Law of Cooling states that the temperature T of an object changes at a rate proportional to the difference between the temperatures of the object ( T ) and the

surrounding medium ( Ts ). That is, k ( T T ) T ( 0 ) T 0 dt

dT = − s =. Using this law, solve the following

problems: A. An object takes 40 minutes to cool from 30o^ C to 24 o^ C in a room having the room- temperature of 20 o^ C. Find the temperature of the object at any time t. What will be the temperature of the object after 15 minutes? How long will it take the object to cool down to 21 o^ C?

Hint: the IVP is = k ( T − 20 ), T ( 0 )= 30 , T ( 40 )= 24 dt

dT ,

Ans.

20 10 40 , ( 15 ) 27. 10 Candwhen 21 ,thetime 100. 5 minutes

ln( 0. 4 ) = + = = =

⎜⎝⎛ ⎟⎠⎞ T e T T t

t

B. One liter of ice-cream at a temperature of –15o^ C is removed from the deep freezer and placed in a room where the temperature is 20o^ C. If after 20 minutes the temperature of the ice-cream is –10o^ C, how long will it take the ice-cream to reach a temperature of 0o^ C?

Ans. 20 35 20 , andwhen 0 ,thetime 72. 63 minutes

ln 6 ln 7 = − = =

⎜⎝⎛^ − ⎟⎠⎞ T e T t

t

C. A thermometer is taken outside from an air-conditioned room where the temperature is 21 o^ C. After 1 minute it reads 27o^ C and after 2 min. it reads 30 o^ C. (That is, T (1) = 27 and T (2) = 30). Find the outside temperature Ts. Ans. Ts = 33o^ C

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Applied Differential Equations (MT2043) Assignment 04

Submission Date: 18/05/

Q5. Find the orthogonal trajectories of all the parabolas with vertices at the origin and foci on the x -axis. ( Hint: here the give family of curves is y 2 = 4 ax ) Ans. 2 x^2 + y^2 = C.

Q6. Find the orthogonal trajectories for the one-parameter family y = – x – 1 + C 1 e x.

Q7. Find a second order homogenous linear differential equation for which the functions

y (^) 1 = x^2 and y 2 = x^2 ln x are solutions. Hint: first find the Wronskian and use it to verify that the given solutions are linearly independent, and then write down the general solution of the required differential equation. Find the required differential equation by eliminating the arbitrary constants from the general solution.

Q8. Use the method of undetermined coefficients to solve the following differential equations:

i.^212432 2

y ′′= 9 x + 2 x − 1 Ans. y = cx + c + x + x − x

ii. (^) y y y e^4 x^ y c 1 e^3 x c 2 e x e^4 x 5

′′ (^) − 2 ′− 3 = 2 Ans. = + − +

iii. y y y e^3 x^ y c 1 e^3 x c 2 e x xe^3 x 2

′′ − 2 ′− 3 = 2 Ans. = + − +

iv. y y y ex^ x y cex cex ex 2 sin x cos x 2

′′ (^) − 2 ′− 3 = 2 − 10 sin Ans. = 1 3 + 2 − + + −

v. (^) x x x x x

x x x

y ce ce x x e x e xe

y y y x e xe e

2 3 2

Ans. 3

2 2 3 2 1 2

2 3

Q9. Use the method of variation of parameters to solve the following differential equations:

i. y cx c xe xe xe x cx cxe xe x x

e y y y x x x x x

x ′′− 2 ′+ = Ans. = 1 + 2 − + ln = 1 + 3 + ln

ii. y y x y c x c x x sin 2 x 12

cos 2 6

′′− 4 =sin^2 2 Ans. = 1 cos 2 + 2 sin 2 +^2 +^2

iii. y ′′+ y =tan x Ans. y = c 1 cos x + c 2 sin x −cos x ( lnsec x +tan x )

iv. (^) 2 4 2 1 2

2 1 2

2 2 2

Ans. ( 1 ) 3

( 1 ) 2 2 6 ( 1 ) given that ( 1 ) y cx c x x x

x y xy y x yc cx c x = + − + +

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