TEXAS A&M UNIVERSITY
DEPARTMENT OF MATHEMATICS
MATH 308-504
Final exam A, 14 Dec 2004
Mark:
Name:
/60
Notes: a) In questions 1-4 you must show working, Maple commands don’t count (that
shouldn’t preclude you from using Maple to check your calculations).
b) In questions 5-6 you may use Maple to solve the equations.
1.
Solve the initial value problem
dy
= −y 2 cos(x)
dx
y(π) = 1.
(6 marks)
2.
Solve the equation
y ′′ + 3y ′ + 2y = e−t
y(0) = 0, y ′ (0) = 2.
Do not use Laplace method here!
(10 marks)
3.
Now solve the same equation,
y ′′ + 3y ′ + 2y = e−t
y(0) = 0, y ′ (0) = 2,
using Laplace transform.
(12 marks)
4.
Solve the initial value problem
(
x′ = x − 5y + 5et ,
y ′ = x − 3y,
x(0) = 5,
y(0) = 1.
(10 marks)
R
I
C
Es
L
5.
In the above circuit C = 1/5F , R = 2Ω, L = 1H and Es = sin(ωt) V . (Remember,
ER = RI, EC = C1 q, EL = LI ′ and q ′ = I)
(a) Write the differential equation satisfied by the current I
(4 marks)
(b) Find the steady state current when ω = 1.
(4 marks)
(c) Find the amplitude of the steady state current when ω =
√
3.
(4 marks)
6.
Cheburashka is diving off a 5 meter platform. His mass is 10kg and he jumps off the
platform with upward velocity of 2m/s. Because of Cheburashka’s big ears, the force due
to air resistance is proportional to his velocity with the proportionality constant 20N-sec/m.
The only other force acting is gravity (g can be approximated by 10m/sec2 ).
(a) Write the differential equation for the velocity v(t).
(2 marks)
(b) Solve the above differential equation and hence find the formula for the position s(t)
of Cheburashka at time t. (Remember, the platform is 5 meters above the water)
(4 marks)
(c) When will Cheburashka hit the water?
(2 marks)
(d) How high above the water will Cheburashka be when at the peak of his trajectory?
(2 marks)
Elementary Laplace transforms
f (t) = L−1 {F (s)}
F (s) = L{f (t)}
1
1
s
s>0
eat
1
s−a
s>a
tn
n!
sn+1
s>0
sin bt
b
s2 +b2
s>0
cos bt
s
s2 +b2
s>0
eat sin bt
b
(s−a)2 +b2
s>a
eat cos bt
s−a
(s−a)2 +b2
s>a
tn eat
n!
(s−a)n+1
s>a
u(t − c)f (t − c)
e−cs F (s)
u(t − c)f (t)
e−cs L {f (t + c)}
ect f (t)
F (s − c)
f ′ (t)
sF (s) − f (0)
f ′′ (t)
s2 F (s) − sf (0) − f ′ (0)
f (n) (t)
sn F (s) − sn−1 f (0) − ... − f (n−1) (0)
(−t)n f (t)
F (n) (s)