ps1sols - 1 • 2 The function G of the hint satisfies the...

Answers to Selected Problems in Chapter 1
1. Problem Set 1.2.1
• 5) This is calculus! x ( t ) = (1 / 3) t 3 - e t + 3.
• 6) x ( t ) = 100 e 0 . 05 t . At t = 1 this gives 105 . 13, as compared with
105 . 12 for monthly compounding.
• 10) If ˙ x ≡ 0 then x = γ/α . The solution for Problem 7 for
arbitrary initial data x (0) = k is
x ( t ) = ( k - γ/α ) e
- αt
+ γ/α,
which tends to γ/α as t → ∞ since α > 0.
• 12)
∂
∂s
f ( s + t ) =
∂
∂t
f ( s + t ) = f 0 ( s + t ), so the partial-differential
equation reduces to 2 f
0
= αf . The solution is u ( s, t ) = C exp((1 / 2) α ( s +
t )) where C is an arbitrary constant.
• 14) The solution is x ( t ) = x
0
exp (sin t ); it is periodic.
• 20) Set F ( u, v ) =
R
u
t
0
g ( v, s ) ds . If u = u ( t ) and v = v ( t ) and
f ( t ) = F ( u ( t ) , v ( t )), the chain rule gives
f
0
( t ) =
∂F
∂u
u
0
( t ) +
∂F
∂v
v
0
( t ) .
Since f ( t ) = F ( t, t ) we put u ( t ) = v ( t ) = t , and obtain the stated
result.
2. Problem Set 1.3.1
• 1) If ˙ x = k
0
x
1+
then
x
- (1+ )
dx
dt
=
d
dt
- x
-
/ = k
0
,
so x - = C - k
0
t where C is a constant. Evaluating this gives
x ( t ) = x
-
0
- k
0
t
- (1 / )
,
which tends to infinity as t tends to t
*
= x
-
0
/ k
0
.
1