Download Math 244 Quiz 4, Solutions 1. a) Find a basis T for R 3 that

Math 244 Quiz 4, Solutions
1.
a) Find a basis T for R3 that contains the vectors v1 = h3, 2, −1i and v2 = h2, −2, 1i .
Apply the method of Example 5 (Section 4.5) to the vectors v1 , v2 , e1 , e2 , and e3 . That is, apply EROs
to the matrix [v1 v2 e1 e2 e3 ] until it is clear which of the three standard basis vectors are independent
of the two vectors v1 and v2 .




3
2 1 0 0
3 2 1 0 0
2R3 +R2
 2 −2 0 1 0  −−−−−→  0 0 0 1 2 
−1
1 0 0 1
−1 1 0 0 1
In the second matrix columns 4 and 5 are independent of columns 1 and 2. The same is true in the first
matrix so columns 1, 2, and 4 of the first matrix are a basis for R3 . Let T = {v1 , v2 , e2 }. Another
solution is T = {v1 , v2 , e3 }.
b) Let A be a 3 × 5 matrix whose three row vectors are linearly independent.
i. What is the dimension of the null space of the matrix A?
ans 5 − 3 = 2
3
ii. Explain why, for each b in R , the nonhomogeneous system Ax = b has a solution.
The matrix A has rank 3 so its column space is 3 dimensional. Since the column space is a subspace
of R3 , it must be all of R3 . Therefore, every vector b in R3 is in the column space of A.
2. With yp = 1 and yc = c1 cos(x) + c2 sin(x), find the solution of y 00 + y = 1 that satisfies the initial conditions
y(0) = −1 = y 0 (0) .
The general solution is y = yc + yp = c1 cos(x) + c2 sin(x) + 1. Observe that y 0 = −c1 sin(x) + c2 cos(x). The
initial condition y(0) = −1 implies that −1 = c1 + 1 so c1 = −2. The condition that y 0 (0) = −1 implies that
−1 = c2 and the solution is y = −2 cos(x) − sin(x) + 1.
3. Obtain the general solution of the differential equation y 00 − 6y 0 + 13y = 0 .
√
6 ± 36 − 52
2
= 3 ± 2i. Therefore, the general
The characteristic polynomial is r − 6r + 13 with roots r1,2 =
2
3x
solution to the differential equation is y = e (c1 cos 2x + c2 sin 2x).
4. (Do not show your work for this problem.) The mass-spring system 25x00 + 10x0 + 226x = 0 is set into motion
with x(0) = 20 and x0 (0) = 41.
a) The system is
over damped critically damped
under damped . (circle one).
b) The position function is x(t) = e−t/5 (20 cos 3t + 15 sin 3t) = 25 e−t/5 cos(3t − φ)
c) The pseudoperiod of the oscillations is
of the envelope curves are x1 (t) =
2π/3 ≈ 2.094
25 e−t/5
.
(3 decimal accuracy). The equations
and x2 (t) =
−25 e−t/5
.
5. Set up the appropriate form of a particular solution yp for the differential equation
y 000 − y 00 = ex + 2x2 − 5 .
Do not determine the values of the coefficients. Hint. The operator D3 (D − 1) annihilates the forcing function.
The particular solution must also be a solution to D5 (D − 1)2 (y) = 0. The characteristic polynomial for this
equation is r5 (r − 1)2 with roots r1 = 0 (5 times) and r2 = 1 (twice) so the particular solution will have the
form
y = c1 + c2 x + c3 x2 + c4 x3 + c5 x4 + c6 ex + c7 xex .
(?)
However, the original homogeneous equation: D2 (D − 1)(y) = 0, has the general solution yc = c1 + c2 x + c6 ex
so this part of (?) should be deleted leaving y = c3 x2 + c4 x3 + c5 x4 + c7 xex . Therefore, the appropriate form
for a particular solution for the original equation is
yp = Ax2 + Bx3 + Cx4 + Dxex .