# Higher Order Differential Equations-Practical Differential Equations-Assignment, Exercises for Applied Differential Equations. B R Ambedkar National Institute of Technology

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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, Posi...
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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)( −=−= πytty

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)( 2772

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

1 72 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()( TTTTkdt 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 040 )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: 18/05/2012

Instructor: Dr. Rashid Ali

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 y2 = 4ax) Ans. 2x2 + y2 = C. Q6. Find the orthogonal trajectories for the one-parameter family y = – x – 1 + C1ex. Q7. Find a second order homogenous linear differential equation for which the functions

xxyxy ln and 22 2

1 == 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. 234212 2 1

3 1

4 3 Ans.129 xxxcxcyxxy −+++=−+=′′

ii. xxxx eececyeyyy 42314 5 2 Ans.232 ++==−′−′′ −

iii. xxxx xeececyeyyy 32313 2 1 Ans.232 ++==−′−′′ −

iv. xxeececyxeyyy xxxx cossin2 2 1 Ans.sin10232 2

3 1 −+++=−=−′−′′

v. xxxxx

xxx

xeexexxececy

exeexyyy

32 2 73 Ans.

42223

2322 21

32

−−+++++=

+++=−′−′′

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

i. xxexecxcxxexexecxcy x

eyyy xxxxx x

lnln Ans.2 3121 ++=+−+==+′−′′

ii. xxxcxcyxyy 2sin 12 12cos

6 12sin2cos Ans.2sin4 2221

2 +++==−′′

iii. ( )xxxxcxcyxyy tanseclncossincos Ans.tan 21 +−+==+′′

iv. 242

21

2 21

222

3)1( Ans.

)1( given that )1(622)1(

xxxcxcy

xcxcyxyyxyx c ++−+=

−+=+=+′−′′+

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