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Math 285 Spring 2003 — Test 2 Solutions
Total points: 100. Do all questions. Explain all answers. No notes, books, calculators or computers.
- [6 points] For the following differential equation, write down the form of the complementary solution yc, and of the particular solution yp (by the method of undetermined coefficients). You do not have to evaluate any coefficients.
y′′^ + 9y = x cos 3x.
yc = c 1 cos 3x + c 2 sin 3x
Here f (x) = x cos(3x) = (polynomial of degree 1) · cos(3x). So we guess yp = (A + Bx) cos 3x + (C + Dx) sin 3x by Undetermined Coefficients Rule 1. But now there is duplication with yc, and so we have to use Rule 2: multiply through by x to get
yp = x(A + Bx) cos 3x + x(C + Dx) sin 3x
Aside. The answer is
yp =
x cos(3x) +
18 x^2 − 1 216
sin(3x),
but you were not required to find this.
- [25=20+5 points] (a) For each value of ω 0 > 0, solve the forced undamped oscillator equation
x′′(t) + ω^20 x(t) = sin(2t),
by Undetermined Coefficients.
Solution. First we write down the complementary solution, solving x′′(t)+ ω 02 x(t) = 0: xc(t) = c 1 cos ω 0 t + c 2 sin ω 0 t.
Then by Undetermined Coefficients we guess xp(t) = A cos 2t + B sin 2t.
There is no duplication if ω 0 6 = 2 , and so in that case we can just sub- stitute xp into the DE to get
[− 4 A cos 2t − 4 B sin 2t] + ω^20 [A cos 2t + B sin 2t] = sin(2t).
Equating coefficients of cos 2t gives − 4 A + ω^20 A = 0 and so A = 0. Equating coefficients of sin 2t gives − 4 B + ω^20 B = 1 and so B = 1/(ω^20 − 4), so that
x(t) = xc(t) + xp(t) = c 1 cos ω 0 t + c 2 sin ω 0 t +
ω^20 − 4
sin 2t.
But there is duplication if ω 0 = 2 , and so in that case we must multiply our guess for xp by t to get xp(t) = t[A cos 2t + B sin 2t]. Substituting this into the DE yields
t′′[A cos 2t + B sin 2t] + 2t′[A cos 2t + B sin 2t]′^ + t[A cos 2t + B sin 2t]′′
- ω^20 t[A cos 2t + B sin 2t] = sin 2t,
where we have used the product rule (F G)′′^ = F ′′G + 2F ′G′^ + F G′′. Taking the derivatives and using ω 0 = 2 gives
2[− 2 A sin 2t+2B cos 2t]+t[− 4 A cos 2t− 4 B sin 2t]+4t[A cos 2t+B sin 2t] = sin 2t,
which simplifies to
− 4 A sin 2t + 4B cos 2t = sin 2t,
so that − 4 A = 1 and 4B = 0, or A = − 1 /4 and B = 0. That is,
x(t) = xc(t) + xp(t) = c 1 cos 2t + c 2 sin 2t −
t cos 2t.
(b) Roughly sketch a typical solution x(t) for large t-values, for ω 0 = 2, 3.
Solution.
- [12 points] A certain shock absorber is described by
mx′′(t) + x′(t) + x(t) = 0.
Find all m-values such that the amplitude of oscillation gets reduced by at least 80%, during each unit of time. (You may assume m > 1 /4.)
Solution. The system is underdamped because c^2 − 4 km = 1^2 − 4 m < 0, using the assumption m > 1 /4. So the solution has the form
x(t) = Ce−pt^ cos(ω 1 t − γ),
where p = c/ 2 m = 1/ 2 m. The amplitude at time t is Ce−pt^ and the amplitude at time t + 1 is Ce−p(t+1). So we want
Ce−p(t+1)^ ≤ (20%)Ce−pt
e−p^ ≤
ep^ ≥ 5 p ≥ log 5 1 2 m
≥ log 5 1 4
< m ≤
2 log 5
- [14=6+8 points] Consider the damped, forced oscillator equation
mx′′(t) + cx′(t) + kx(t) = F 0 cos(ωt),
where m, c, k are positive constants.
(a) Write down the form of the steady periodic response to the forcing. (You do not have to evaluate any of the coefficients.)
Solution. xp(t) = A cos(ωt) + B sin(ωt) by Undetermined Coefficients.
There is definitely no duplication with xc(t): damping is present (c > 0) and so the complementary solution involves a decaying exponential (xc is transient), whereas our guess for xp has no exponentials.
(b) Take x(0) = 1000, x′(0) = 2000 and ω = 2. Roughly sketch the shape of the solution x(t) for large t, as best you can. Explain.
Solution. x = xc + xp and the complementary solution is transient due
to the damping, xc(t) → 0 as t → ∞. So xc is negligible for large t. Hence the initial conditions are irrelevant for large t (since they only affect xc and not xp). What matters is just the frequency and amplitude of the response xp to the forcing. The forcing frequency is ω = 2 , and the amplitude of xp (in other words A and B) cannot be found by Undetermined Coefficients because we are not given values for m, c, k, F 0. So the best graph we can draw for large t is a simple oscillation with frequency 2, period π:
Warning. A and B have nothing to do with the initial conditions 1000 and 2000. The values of A and B can only be determined by substituting xp into the original DE (Method of Undetermined Coefficients).
The boundary condition X′(0) = 0 implies B = 0, since cos(0) = 1 and sin(0) = 0 and λ 6 = 0 by assumption. So now X(x) = A cos(
λx). Then the boundary condition X′(π) = 0 implies
−A
λ sin(
λπ) = 0.
Now, λ 6 = 0 by assumption, and so either A = 0 or else sin(
λπ) = 0. If A = 0 then X ≡ 0 again. So our only chance for λ to be an eigenvalue is if sin(
λπ) = 0. This happens whenever
λ is a positive integer, or in other words: λ = 1^2 , 22 , 32 ,....
So these λ-values (and no others) are the positive eigenvalues. We saw above that the eigenfunction corresponding to eigenvalue λ > 0 is X(x) = A cos(
λx). By taking A = 1 for convenience we get the following eigenfunctions corresponding to λ = 1^2 , 22 , 32 ,.. .:
X(x) = cos(1x), cos(2x), cos(3x),....
Remembering from above that λ = 0 is also an eigenvalue, with eigen- function X = 1, we have shown that the complete collection of eigenvalues
is λn = n^2 for integers n = 0, 1 , 2 , 3 ,... , with corresponding eigenfunctions
Xn(x) = cos(nx). (Check. When n = 0, this reduces back to λ 0 = 0 and X 0 (x) = 1.)
(b) If n, m ≥ 0 and n 6 = m then
∫ (^) π 0 cos(nx) cos(mx)^ dx^ = 0.^ Explain why this is true.
Solution. By part (a) we know Xn(x) = cos(nx) and Xm(x) = cos(mx) are eigenfunctions satisfying homogeneous Neumann boundary conditions at x = 0, π, with eigenvalues λn = n^2 and λm = m^2. If n 6 = m then λn 6 = λm, and so Xn and Xm are orthogonal by the Orthogonality of Eigenfunctions Theorem. That is,
∫ (^) π
0
Xn(x)Xm(x) dx =
∫ (^) π
0
cos(nx) cos(mx) dx.
Formulas Here are some formulas you might be able to use on the test:
y = yc + yp
ω 0 =
k m
, p =
c 2 m
, ω 1 =
ω^20 − p^2
e(a±ib)x^ = eax(cos bx ± i sin bx)
y = −y 1
y 2 f W
dx + y 2
y 1 f W
dx
W = y 1 y′ 2 − y′ 1 y 2
