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Name:_____________________________
Math 251H Section 001 – Quiz #6
1. (a) Find the general solution of
y(4) + y0 = 0.
(b) How many initial conditions would you need for an initial value problem with this equation?
(c) How can you confirm that the different solutions are linearly independent? Do not show.
Solutions:
(b) 4th order ODE so require 4 initial conditions.
(c) Compute the Wronskian again!
(a) Let y = ert and recover the characteristic equation
r4 + r = 0
r(r3 + 1) = 0
which has roots r = 0, so y1 = C1 solves the equation, and
r3 = −1
= exp (πi + 2nπi)
π 2nπ
⇒ r = exp
+
i
3
3
n = 0, 1, 2. Then
π 2nπ
π 2nπ
+
+ i sin
+
r = cos
3
3
3
3
√

3
1

2 + i 2 , n = 0
=
−1, √
n=1

1
3
n = 2,
2 −i 2
√
√
and therefore y2 = C2 e−t , y3 = C3 et/2 cos( 3t/2), y4 = C4 et/2 sin( 3t/2).
The general solution of y(4) + y0 = 0 is
√
√
y(t) = C1 +C2 e−t +C3 et/2 cos( 3t/2) +C4 et/2 sin( 3t/2).
2. An external force F(t) = 2 cos 2t is applied to a mass–spring system with m = 1 kg, damping coefficient
b = 0, and spring constant k = 4, which is initially at rest.
(a) Write down an initial value problem for that describes the distance of the mass from equilibrium, x(t).
(b) Without solving: what will eventually (as t increases) happen to the spring? Why?
Solution: (a) The IVP for this system is
(
x00 + 4x = 2 cos 2t
x(0) = x0 (0) = 0.
(b) Note that the natural frequency ω = 2 matches the forcing frequency β = 2. This is our resonant case.
Since there is no damping (b = 0), the solution will therefore show that the amplitude of oscillation grows
infitely large. However, the spring isn’t infinitely long! So the spring will eventually break.
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3. A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The
initial charge is zero. If a 12-volt battery is connected to the circuit and the circuit is closed at t = 0,
determine the limiting charge as t → ∞.
Solution: Since L = 1 H, R = 5 × 103 Ω, C = 0.25 × 10−6 F, impressed voltage E(t) = 12 V, and the
initial charge 0, the initial value problem for the charge in the LCR circuit is
(
Q00 + 5 × 103 Q0 + 4 × 106 Q = 12
Q(0) = Q0 (0) = 0.
SIMPLE APPROACH:
There’s resistance (i.e. damping) on the circuit. We know that therefore, the homogeneous solution Qh (t)
will → 0 as t → ∞ (could be overdamped, damped oscillations, doesn’t matter). For long-time behavior
we only need check the particular solution.
Use the method of undetermined coefficients. Let Q p = A. Since Q0p = Q00p = 0, plugging in to the ODE
we recover
4 × 106 A = 12
A = 3 × 10−6 .
Therefore Q p = 3 × 10−6 . Taking the limit,
lim Q(t) = lim Qh (t) + lim Q p (t)
t→∞
t→∞
t→∞
−6
= (0) + 3 × 10
= 3 × 10−6 .
The limiting charge is 3 × 10−6 C.
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HARDER APPROACH – SOLVING THE FULL SYSTEM
First, the homogeneous solution.
The characteristic equation associated with the homogeneous ODE is
r2 + 5 × 103 r + 4 × 106 = 0
which has roots
p
(5 × 103 )2 − 4 (4 × 106 )
r =
2
= −1000, −4000.
−5 × 103 ±
Therefore the homogeneous solution is Qh = C1 e−1000t +C2 e−4000t .
For the particular solution, use the method of undetermined coefficients. Let Q p = A. Since Q0p = Q00p = 0,
plugging in to the ODE we recover
4 × 106 A = 12
A = 3 × 10−6 .
Therefore Q p = 3 × 10−6 .
The general solution for the charge is
Q(t) = C1 e−1000t +C2 e−4000t + 3 × 10−6 .
COULD STOP HERE AND GET THE SOLUTION, TOO:
Regardless of C1 and C2 – which we can find from the initial conditions – it’s clear that the limiting charge
is
lim Q(t) = 3 × 10−6 C.
t→∞
OR FINDING THE CONSTANTS...
Since Q(0) = 0 and Q0 (0) = 0,
C1 +C2 = −3 × 10−6
−1000C1 − 4000C2 = 0.
Solving simultaneously,
C1 = 4 × 10−6 , C2 = −1 × 10−6 .
Therefore the charge on the capacitor at time t is
Q(t) = 4 × 10−6 e−1000t − 1 × 10−6 e−4000t + 3 × 10−6 C.
And the limiting charge as t → ∞ is
lim Q(t) = 3 × 10−6 C.
t→∞
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