MA1506
Tutorial 9
SOLUTIONS
Question 1. The thing to remember here is that there are six possible states in which
the cunning Miss Tan can nd herself: she can have $0, $1, $2.up to $5. She can never
have more than that because Ah Huat wont allow it. If at
MA1506 TUTORIAL 8 SOLUTIONS
Question 1
From the given hint, we see that we have to solve
d4 y
Mg
(x A),
=
4
dx
EI
subject to the given boundary conditions. [Note that y (0) = y (0) = 0 since the pole is
horizontal at the point where it joins the wall. In
ANSWERS TO MA1506 TUTORIAL 7
Question 1.
(a) We shall use the following s-Shifting property:
L(f (t) = F (s) L(ect f (t) = F (s c)
n!
2
n
2
L(t ) = 3 use L(t ) = n+1
s
s
2
2 3t
) = L(e3t t2 ) =
L(t e
(s + 3)3
(b) Here u denotes the Unit Step Function gi
ANSWERS TO MA1506 TUTORIAL 6
Question1.
B2
.
4s
From Tutorial 5 we know B = 1.5 and and N = 376, so N = B/s s =
First compare 80 with
2
B
4s
1.5
376
= 141. This is the maximum number we can kill without causing extinction.
Setting E = 80,
B
1
=
2
B 2 4Es
ANSWERS TO MA1506 TUTORIAL 5
Question 1
The cost of making the can is a function of the radius r given by
2V
) + K 2r
r
2V
C (r) = J (4r 2 ) + 2K
r
,
= J (4r 2h) + 2K = 0
C (r) = J (2r2 +
h
K/J
=2+
r
r
where J is the cost of aluminium per square cm, V is
MA1506 TUTORIAL 4 SOLUTIONS
Question 1
(i) x = cosh(x). An equilibrium solution of an ODE is just a solution that is identically
constant. That is not possible here because the cosh function never vanishes. So there is
no equilibrium for this ODE.
(ii) x
MA1506 Tutorial 3 Solutions
(1a)
y"+6 y '+9 y = 0
Set y = e t
2 + 6 + 9 = 0 = 3
y = ( A + Bx)e 3 x y ' = Be 3 x 3( A + Bx)e 3 x
y (0) = 1 A = 1
y ' (0 ) = 1 B 3 A = 1 B = 2 y = (1 + 2 x)e 3 x
(1b)
2 2 + (1 + 4 2 ) = 0 = 1 2i
y = e x [ A cos 2x + B sin 2
MA1506 Tutorial 2 Solutions
Question 1.
(1a)
1
1
y '+(1 + ) y = e x
x
x
1
Integrating factor is exp (1 + ) = exp( x + ln x) = xe x (in general, y '+ P ( x) y multiply
x
d
( ye P ) )
by exp P y ' e P + Pye P =
dx
d
1
( yxe x ) = 1(= xe x e x )
dx
x
x
x
yx
MA1506 Tutorial 1 Solutions
(1a)
y' =
1
1
1
x
=
y = ln
+c
x +1
x( x + 1) x x + 1
(1b)
y ' = cos x cos 5 x =
1
[cos 6 x + cos 4 x] y = 1 1 sin 6 x + 1 sin 4 x + c
2
2 6
4
(1c)
dy
1
= e x e 3 y e 3 y dy = e x dx e 3 y = e x + c
dx
3
(1d)
1+ y
1
dy = (2 x 1