Mathematics MATB44, Assignment 2
Solutions to Selected Problems
Question 1.
Solve
4y 00 − 12y 0 + 9y = 0
Soln: The characteristic equation is
4r2 − 12r + 9 = 0
The solutions are
r=
24 ±
√
144 − 16 × 9
=3
8
(repeated root)
So the solutions are
y = e3x
and
y = xe3x
Question 2
y 00 − 6y 0 + 25y = 0;
y(0) = 3, y 0 (0) = 1
Soln: The characteristic equation is
r2 − 6r + 25 = 0
r=
√
1
6 ± 36 − 100 = 3 ± 4i
2
Solutions are
y = c1 e3x cos(4x) + c2 e3x sin(4x)
y 0 (x) = c1 (3e3x cos(4x) − 4e3x sin(3x)) + c2 (3e3x sin(4x) + 4e3x cos(4x))
y(0) = c1 = 3
y 0 (0) = 3c1 + 4c2 = 1
9 + 4c2 = 1
1
c2 = −2
Soln
y = 3e3x cos(4x) − 2e3x sin(4x)
Question 3. Solve y 00 + k 2 y = 0. The characteristic equation is
r2 + k2 = 0
whose solutions are
r = ±ki
so the solution is
y(x) = C1 cos(kx) + C2 sin(kx).
Question 4. Find an ODE with solutions (a) 2ex , e−5x
These functions are solutions of the equation
y 00 + ay 0 + by = 0
if the characteristic equation is r2 + ar + b = 0 with solutions r = 1 and
r = −5, in other words (r − 1)(r + 5) = 0 or
r2 + 4r − 5 = 0.
So the equation is
y 00 + 4y 0 − 5y = 0.
(b) e−x cos(x), e−x sin(x) These functions solve the ODE with characteristic equation whose roots are −1 ± i, in other words the characteristic
equation is
(r + 1 + i)(r + 1 − i) = 0
or
r2 + 2r + 2 = 0.
Question 5.
xy 00 − 2(x + 1)y 0 + 4y = 0; y1 (x) = e2x
We seek a solution of the form y2 = vy1 where
(v 0 )0 + (p + 2(y1 )0 /y1 )v 0 = 0
2
So
0
−2 −
v = (y1 ) e
R
p
= e−4x e2x e2lnx
= e−2x x2
So
v=
Z
e−2x x2 dx
Integration by parts:
Z
e
−2x 2
x dx =
Z
udw = uw −
Z
wdu
where dw = e−2x dx and u = x2 so w = − 21 e−2x and du = 2xdx. So
Z
Z
1
e−2x x2 dx = x2 e−2x + xe−2x dx
2
Z
xe−2x dx =
Z
udw = uw −
Z
wdu
where dw = e−2x dx and u = x so w = − 12 e−2x and du = dx so
Z
So
xe
−2x
1
1
1 −2x 1 Z −2x
+
e dx = − xe−2x − e−2x .
dx = − xe
2
2
2
4
1
1
1
x2 e−2x dx = − x2 e−2x − xe−2x − e−2x .
2
2
4
So the solution is
Z
1
1
1
y2 = vy1 = (− x2 − x − .
2
2
4
In fact any constant multiple of this is also a solution: so
y 2 = x2 + x +
1
2
is a solution. Question 6.
(x2 + 1)y 00 − 2xy 0 + 2y = 0; y1 (x) = x
y 00 + p(x)y 0 + q(x)y = 0
3
where
−2x
.
x2 + 1
p(x) =
The solution is y2 (x) = v(x)y1 where
v 0 = y1−2 e−
= x−2 e
R
R
p(x)dx
2x
dx
x2 +1
.
Substitute w = x2 , dw = 2xdx or
Z
Z
2x
dw
dx =
2
x +1
w+1
= ln |w + 1| = ln(x2 + 1). So
1 ln(x2 +1) x2 + 1
v = 2e
=
.
x
x2
0
v0 = 1 +
1
,
x2
or
v = x − 1/x.
Thus
y2 = vy1 = (x − 1/x)x = x2 − 1.
Question 7.
x2 y 00 − 2xy 0 + (x2 + 2)y = 0; y1 = x cos(x).
We seek a solution of the form y2 = vy1 where v 0 = y1−2 e−
−2x/x2 . Here
R
R
e− p(x)dx = e 2/xdx = e2 ln |x| = x2 .
So
v0 =
and
v=
x2
Z
1
x2 = sec2 (x),
cos(x)
sec2 (x)dx = tan(x).
Thus
y2 = x tan(x) cos(x) = x sin(x).
4
R
pdx
for p(x) =
Question 8.
xy 00 − y 0 + 4x3 y = 0
y1 (x) = cos(x2 ).
The equation is
y 00 + p(x)y 0 + q(x)y = 0
where p(x) = −1/x so − pdx = ln |x|.
R
y2 = vy1
where v 0 (x) = y1−2 e−
R
pdx
=
x
.
cos2 (x2 )
v(x) =
=
where u = x2
=
Z
x sec2 (x2 )dx
1Z
du sec2 (u)
2
1
1
tan(u) = tan(x2 ).
2
2
So the solution is
y2 =
1
1
tan(x2 ) cos(x2 ) = sin(x2 ).
2
2
Question 9.
x2 y 00 + xy 0 − 9y = 0; y1 = x3
9
1
y 00 + y 0 − 2 y = 0
x
x
y2 = vy1
where v satisfies
v 0 = y1−2 e−
R
pdx
= x−6 e− ln(x) = x−7 .
So
and
1
v = − x−6
6
1
y2 = − x−3 .
6
5
Thus another solution is y = x−3 .
Question 10.
(1 − x2 )y 00 − 2xy 0 + 2y = 0
y 00 + p(x)y 0 + q(x)y = 0
2x
where p(x) = − 1−x
2 for |x| < 1.
Z
p(x)dx = −
Z
Z
2x
dw
dx = −
2
1−x
1−w
2
where w = x2 . This
R integral equals ln |1 − w| = ln(1 − x ). Hence y2 = vy1
1
where v 0 = y1−2 e− pdx = x−2 1−x
2 So
v=
Z
dx
dx
=
x2 (1 − x2 )
x2 (1 + x)(1 − x)
Z
Now we use partial fractions:
1
A
B
A − Ax2 + Bx2
=
+
=
.
x2 (1 − x2 )
x2 1 − x2
x2 (1 − x2
It follows that A = B = 1, so
x2 (1
Now
1
1
1
= 2 ++
2
−x )
x
1 − x2
1
C
D
=
+
2
1−x
1−x 1+x
C(1 + x) + D(1 − x)
=
(1 − x)(1 + x)
Hence C = D = 1/2 and
1
1
1
=
+
.
(1 − x)(1 + x)
2(1 − x) 2(1 + x)
So
1
1
1
1
=
+
+
.
x2 (1 − x)(1 + x)
x2 1 − x 1 + x
6
So
1
v = − − ln |1 − x| + ln |1 + x|
x
1 + x
1
.
= − + ln x
1 − x
Question 11. (a) f (x) = x3 , g(x) = x2 |x| so g(x) = x3 if x > 0 but
g(x) = −x3 if x < 0.
W (f, g) = f g 0 − gf 0
W (f, g) = 0 if x < 0 since for x < 0 f = −g
W (f, g) = 0 if x > 0 since for x > 0 f = g
For x = 0, f = f 0 = g = g 0 = 0 so W (f, g) = 0 at x = 0.
(b) If f and g were linearly dependent, there would be C1 , C2 with
C1 f (x) + C2 g(x) = 0
for all x. But then
C1 x3 + C2 x2 |x| = 0
C1 + C2 x/|x| = 0
so C1 = −C2 and C1 = C2 . Hence C1 = C2 = 0.
(c) Theorem 3.3.3 (text, 8th edition) reads:
Let y1 and y2 be the solutions of
L[y] = y 00 + p(t)y 0 + q(t)y = 0
where p and q are continuous on an open interval I. Then y1 and y2
are linearly dependent on I if and only if W (y1 , y2 )(t) is zero for all t in
I. Alternatively, y1 and y2 are linearly independent on I if and only if
W (y1 , y2 )(t) is never zero in I.
Parts (a) and (b) do not contradict these Theorems since the Theorems
are only valid under the hypothesis that f and g solve some differential
equation L[f ] = 0 and L[g] = 0.
(Similar theorems in the 9th edition are Theorem 3.2.3 and Theorem
3.2.4. These two theorems would also yield a contradiction if f and g solved
L[f ] = 0 and L[g] = 0.)
Question 12.
A torpedo is travelling at a speed of 90 km/hour at the moment it runs
out of fuel. If the water resists its motion with a force proportional to the
7
speed and if 1 km of travel reduces its speed to 60 km/hour, how far will it
travel before it stops?
Soln: Let y(t) be the distance the torpedo has travelled. Let t = 0 be the
time when it runs out of fuel.
So y(0) = 0
dy
(0) = 90km/h
dt
Let the force of the water on the torpedo be −b dy
(proportional to the speed).
dt
So the equation of motion of the torpedo is
m
or
d2 y
dy
+b =0
2
dt
dt
b dy
d dy
( )+
=0
dt dt
m dt
This is a first order equation for
dy
:
dt
Define v(t) =
dy
,
dt
then
dv
b
+ v=0
dt m
dv
b
= − dt
v
m
so
ln|v| = −
b
t + C1
m
b
|v| = C2 e− m t
or
b
v(t) = C3 e− m t .
b
m
Introduce K1 = C3 , K2 =
(two constants which are unknown). Then
v(t) = K1 e−K2 t
y(t) =
Z t
0
=−
0
0
v(t )dt = K1
Z t
0
e−K2 t dt0
0
K1 −K2 t0 t
K1
e
|0 =
(1 − e−K2 t )
K2
K2
1 km of travel reduces the speed to 60 km/h
8
So for T such that y(T ) = 1,
1
(1 − e−K2 T ) = 1,
When K
K2
dy
(T )
dt
= 60 and
dy
(0)
dt
= 90, we get K1 = 90
K1 e−K2 T = 60
in other words
or
K1
3
90
=
=
K1 e−K2 T
60
2
3
2
3
K2 T = ln( )
2
eK2 T =
But
K1
2
1 K1
K1
(1 − e−K2 T ) =
(1 − ) =
=1
K2
K2
3
3 K2
or
K1
= 3.
K2
The distance travelled by the torpedo before stopping is
lim y(t) =
t→∞
K1
=3
K2
So the torpedo travels 3 km before stopping.
Question 13 Solve the difference equation
n+2
yn
n+3
for the initial value y0 . Also describe the behaviour as n → ∞
Soln:
yn+1 =
2
y1 = y0
3
3
2
y2 = y1 = y0
4
4
4
2
y3 = y2 = y0
5
5
Conjecture:
yn =
2
y0
n+2
9
Proof by induction:
True for n = 1 (base case)
2
2
y0 . But yn+1 = n+2
y = n+3
y0 so the
Suppose true for n. Then yn = n+2
n+3 n
statement is also true for n + 1. This completes the proof by induction.
As n → ∞, yn → 0.
Question 14
A student deposits $500 in a bank account. The interest rate is 6 %
compounded monthly. The student also deposits $10 in the account every
month. How much money is in the account after 2 years?
Solution:
The balance in the n-th month is
yn = ρyn−1 + b
where b = 10 (in dollars) and ρ = 1.005 (because 6 percent yearly = 0.5 percent compounded monthly). We also know that a0 = $ 500 (initial deposit).
So the solution is (as in textbook chapter 2.9 equation (14))
yn = ρn y0 +
1 − ρn
b
1−ρ
where ρ = 1.005 and b = 10.
When n = 24 this gives
1.00524 · 500 + 0.27(200 × 10) = 563.60 + 254.32 = 817.92
(in dollars)
Question 14, as originally stated
A student deposits $500 in a bank account. The interest rate is 5%
compounded monthly. The student also deposits $10 in the account every
month. How much money is in the account after 2 years?
Solution:
The balance in the n-th month is
yn = ρyn−1 + b
where b = 10 (in dollars) and ρ = 1 + 0.01 × (5/12) = 1.000416666 (because
5 percent yearly = 5/12 percent compounded monthly). We also know that
a0 = $ 500 (initial deposit).
So the solution is (as in textbook chapter 2.9 equation (14))
10
yn = ρn y0 +
1 − ρn
b
1−ρ
where ρ = 1.0041666666 and b = 10.
When n = 24 this gives
(1.0041666)24 · 500 + 10(0.1049413)/0.0041666 = 552.47 + 265.19 = 817.66
(in dollars)
Question 14 (monthly interest rate 5%)
For those students who thought ’the interest rate is 5 % compounded
monthly’ meant ’the monthly interest rate is 5%’, the answer would be
ρ = 1.05
ρ24 = 3.2251
so
yn = ρn y0 +
1 − ρn
b
1−ρ
(1.05)24 · 500 + 10(2.2251)/0.05 = 1672.55 + 200 × 2.2251 = 2057.57
(in dollars)
11