diffEq HW (sliced finger edition)
Problem 1
For the differential equation
2
y − 2+
t
′′
2
y + 1+
y = et ,
t
′
0 < t < ∞,
with y1 (t) = et given as a solution of the homogeneous equation.
(a) Complementary Solution
Reduction of order. Let y2 = v(t)et . Then
y2′ = (v ′ + v)et ,
y2′′ = (v ′′ + 2v ′ + v)et .
2
2
y′ + 1 +
y = 0 divide by et :
y ′′ − 2 +
t
t
2
2
(v ′ + v) + 1 +
v = 0.
v ′′ + 2v ′ + v − 2 +
t
t
Collect coefficients of v ′′ , v ′ , and v:
• v ′′ coefficient: 1
2
• v ′ coefficient: 2 − 2 + 2t = −
t
2
• v coefficient: 1 − 2 + t + 1 + 2t = 0
This reduces to
2
v ′′ − v ′ = 0.
t
let w = v ′
w′ =
2
w′
2
w→
= → ln (w) = 2 ln (t)
t
w
t
w = t2
t3
let v = t3
3
y2 (t) = t3 et
v ′ = t2 → v =
yC (t) = c1 et + c2 t3 et
(b) Particular Solution via Variation of Parameters
Wronskian.
W (y1 , y2 ) = y1 y2′ − y1′ y2 .
1
With y1′ = et and y2′ = (3t2 + t3 )et :
W = et (3t2 + t3 )et − et · t3 et = 3t2 e2t .
Formula (standard form, so g(t) = et ):
Z
Z
y2 g
y1 g
dt + y2
dt.
W
W
Z
t
y2 g
t3 et · e t
t
t2
=
=
→
dt = ,
2
2t
W
3t e
3
3
6
Z
y1 g
1
1
et · e t
1
=
dt = − .
= 2 2t = 2 →
2
W
3t e
3t
3t
3t
2
2 t
2 t
t
1
t
e
t
e
t2
yp = −et
+ t 3 et −
=−
−
= − et .
6
3t
6
3
2
yp = −y1
General solution.
y(t) = c1 et + c2 t3 et −
t2 t
e
2
Problem 2
A 10-kg mass suspended from a spring stretches it 9.8 cm. At t = 0 the system, initially at rest, is disturbed
by F (t) = 20 cos(8t) N (positive downward). There is no damping.
(a) Spring Constant
At equilibrium, the spring force balances gravity: mg = kL, where L = 0.098 m.
2
(10 kg)(9.8 m/s )
mg
=
= 1000 N/m.
L
0.098 m
k=
(b) Initial Value Problem and Solution
my ′′ + ky = F (t) =⇒ 10y ′′ + 1000y = 20 cos(8t) =⇒ y ′′ + 100y = 2 cos(8t).
Initial conditions (starts at rest):
y ′ (0) = 0.
√
Homogeneous solution. Natural frequency ω0 = 100 = 10 rad/s:
y(0) = 0,
yh (t) = c1 cos(10t) + c2 sin(10t).
Particular solution. Since the forcing frequency (8 rad/s) does not match ω0 = 10 rad/s, try yp =
A cos(8t):
1
−64A + 100A = 2 =⇒ 36A = 2 =⇒ A =
.
18
1
cos(8t).
yp (t) =
18
1
Apply initial conditions to y(t) = c1 cos(10t) + c2 sin(10t) + 18
cos(8t):
y(0) = 0 :
c1 +
1
1
= 0 =⇒ c1 = − .
18
18
8
sin(8t),
18
10c2 = 0 =⇒ c2 = 0.
y ′ (t) = −10c1 sin(10t) + 10c2 cos(10t) −
y ′ (0) = 0 :
2
cos A − cos B = 2 sin
A+B
2
y(t) =
sin
B−A
2
:
1
1
1
cos(8t) −
cos(10t) =
[cos(8t) − cos(10t)] .
18
18
18
y(t) =
1
sin(t) sin(9t)
9
This is a beat phenomenon: a fast oscillation (sin 9t) modulated by a slow envelope (sin t).
(c) Maximum Excursion
Since
1
1
| sin t| | sin 9t| ≤ ,
9
9
the displacement can never exceed 1/9 m. This bound is attained at t = π/2, where sin(π/2) = 1 and
sin(9π/2) = sin(π/2) = 1 simultaneously:
π 1
= .
y
2
9
Checking y ′ (π/2) = 0 confirms this is a critical point, and by the bound above it is the global maximum.
|y(t)| =
ymax =
1
π
m ≈ 0.111 m at t = s ≈ 1.571 s
9
2
Problem 3
The equation y (4) − y ′′ + y = 0 has a fundamental set of solutions with Wronskian W (0) = 1. Find W (4).
Abel’s theorem. For
y (4) + p3 (t)y ′′′ + p2 (t)y ′′ + p1 (t)y ′ + p0 (t)y = 0,
the Wronskian satisfies
Z t
W (t) = W (t0 ) exp −
p3 (s) ds .
t0
Here the equation is
y (4) + 0 · y ′′′ − y ′′ + 0 · y ′ + y = 0,
so p3 (t) = 0 (there is no y ′′′ term). Therefore
W (t) = W (0)e0 = W (0) = 1
for all t.
W (4) = 1
Problem 4
The equation y ′′′ + sin(t)y ′′ + cos(t)y ′ + 2y = 0 has Wronskian W (1) = 0. Find W (t).
By Abel’s theorem, with p2 (t) = sin(t) as the coefficient of y ′′ :
Z t
W (t) = W (1) exp −
sin(s) ds .
1
Since W (1) = 0, and the exponential factor is never zero:
W (t) = 0 for all t
This makes sense: Abel’s formula shows W (t) is W (1) times a nonzero exponential factor, so if W (1) = 0,
the Wronskian is identically zero, the solutions in this fundamental set are actually linearly dependent
everywhere.
3
Problem 5
Solve y (4) − y ′′′ = 0 with y(0) = 0, y ′ (0) = 0, y ′′ (0) = 0, y ′′′ (0) = 1.
(a) General Solution
r4 − r3 = 0 =⇒ r3 (r − 1) = 0.
Roots: r = 0 (multiplicity 3) and r = 1.
y(t) = c1 + c2 t + c3 t2 + c4 et
(b) Initial Value Problem
Compute derivatives:
y ′ (t) = c2 + 2c3 t + c4 et ,
y ′′ (t) = 2c3 + c4 et ,
y ′′′ (t) = c4 et .
Apply initial conditions:
y(0) = 0 :
c1 + c4 = 0
y ′ (0) = 0 :
c2 + c 4 = 0
y ′′ (0) = 0 :
2c3 + c4 = 0
′′′
y (0) = 1 :
c4 = 1
1
c3 = − ,
2
c2 = −1,
Solving from the bottom up:
c4 = 1,
c1 = −1.
1
y(t) = −1 − t − t2 + et
2
Check: y ′′′ (t) = et , y (4) (t) = et , so y (4) − y ′′′ = et − et = 0
Problem 6
Given the general solution
y(t) = c1 e−t sin t + c2 e−t cos t + c3 et sin t + c4 et cos t,
determine a3 , a2 , a1 , a0 in
y (4) + a3 y ′′′ + a2 y ′′ + a1 y ′ + a0 y = 0.
Characteristic roots
−1 ± i
(from e−t sin t, e−t cos t),
1±i
(from et sin t, et cos t).
[r − (−1 + i)][r − (−1 − i)] = (r + 1)2 + 1 = r2 + 2r + 2,
[r − (1 + i)][r − (1 − i)] = (r − 1)2 + 1 = r2 − 2r + 2.
Multiply the two quadratics:
(r2 + 2r + 2)(r2 − 2r + 2) = (r2 + 2)2 − (2r)2 = r4 + 4r2 + 4 − 4r2 = r4 + 4.
4
Characteristic equation is
r4 + 4 = 0.
r 4 + a 3 r 3 + a 2 r 2 + a 1 r + a0 = 0
a3 = 0,
a2 = 0,
a1 = 0,
The differential equation is
y (4) + 4y = 0.
5
a0 = 4