### MATH 111 (FALL 2013) - ASSIGNMENT 1 SELECTED SOLUTION

```MATH 111 (FALL 2013) - ASSIGNMENT 1
SELECTED SOLUTION
PROBLEM 2: (Ch. 2, #6) Consider the harmonic oscillator system x00 + bx0 + kx = 0 where k 6= 0.
(a) Rewrite this second-order equation into a first-order linear system of the form X0 = AX.
(b) Find all values of b and k for which this system has real, distinct eigenvalues, and find the general solution
of this system in these cases. (Use of software such as Mathematica is allowed provided that you indicate
which part was done by the machine.)
1
(c) Find the solution of the system that satisfies the initial condition X(0) =
.
0
(d) Describe the motion of the mass in (c).
Sketch of solutions:
(a) Let v = x0 and X =
x
0
, then X0 =
v
−k
1
X.
−b
(b) Characteristic equation is: λ2 + bλ + k = 0, roots are:
√
√
−b + b2 − 4k
−b − b2 − 4k
λ1 =
, λ2 =
.
2
2
Real distinct if b2 − 4k > 0.
(
0x + v
= λ1 x
To find eigenvectors for λ1 , we solve
. We can avoid lengthy computations by noting
−kx − bv = λ1 v
that one of the equations must be redundant as we expect to get infinitely many solutions. Ignore
the more difficult second equation and we get v = λ1 x. Similarly for λ2 .
Corresponding eigenvectors are:
1
v1 =
,
λ1
1
v2 =
.
λ2
(
c1 + c2
=1
for c1 , c2 . Use whatever method you like but in my opinion
λ1 c1 + λ2 c2 = 0
the Cramer’s Rule works the best here. One should get:
λ2
λ1
c1 = −
, c2 =
.
λ1 − λ2
λ1 − λ2
λ2
λ1
λ1 t 1
λ2 t 1
X(t) = −
e
+
e
λ1
λ2
λ1 − λ2
λ1 − λ2
where λ1 and λ2 are the eigenvalues found in (b).
(c) It amounts to solving
2
1
eλ1 t + λ1λ−λ
eλ2 t . To be physically relevant, you may assume b ≥ 0 and k > 0 (see
(d) x(t) = − λ1λ−λ
2
2
Textbook P. 26, I should have stated that in the homework). From (b) we have λ1 < 0 and λ2 < 0.
Also by observing that λ2 < λ1 , one can check (from elementary calculus) that x(t) > 0. Therefore the
mass will go towards the origin as t increases, and there is no oscillation.
1
PROBLEM 4: (Ch. 2, #11 modified) Prove that two vectors v =
v1
w1
and w =
are linearly
v2
w2
dependent if and only if
v
det 1
v2
w1
= 0.
w2
Solution: (⇒) Suppose v and w are linearly dependent. WLOG, assume v = cw for some real scalar c.
Then we have v1 = cw1 and v2 = cw2 .
v1 w1
det
= v1 w2 − v2 w1 = cw1 w2 − cw2 w1 = 0.
v2 w2
v w1
(⇐) Conversely, suppose 0 = det 1
= v1 w2 − v2 w1 , then v1 w2 = v2 w1 . If both v1 = 0 and v2 = 0
v2 w2
then v = 0 and therefore v = 0w. Now assume either one of {v1 , v2 } is non-zero, and WLOG assume
v1 6= 0, then we have
v2
w2 = w1 .
v1
One can then verify that
w1 v1
w1
w1
=
w = v2
v.
=
v 1 v2
v1
v1 w1
Therefore v and w are linearly dependent.
PROBLEM 5: Consider the following system Y0 (t) = AY(t) + b where A is a 2 × 2 matrix with constant
entries and b ∈ R2 is a fixed vector. Given that Y0 (t) is a particular solution to the system. Prove that any
solution Y(t) to the system Y0 (t) = AY(t) + b can be written in the form:
Y(t) = Y0 (t) + X(t)
0
where X(t) is a solution to the system X (t) = AX(t).
Solution: Given any solution Y(t) to the system Y0 = AY + b, we first verify that Y − Y0 is a solution
to the homogeneous system X0 = AX:
(Y − Y0 )0 = Y0 − Y00
= (AY + b) − (AY0 + b)
since both Y and Y0 are solutions to the b-system.
= AY − AY0
= A(Y − Y0 ).
Now it shows Y − Y0 satisfies the equation X0 = AX. Letting X = Y − Y0 yields
Y = Y0 + X
and from above, this X is a solution to the homogeneous system X0 = AX. It completes the proof.
Remark: This problem tells us that to find the general solution to the system Y0 = AY + b, we just need
to solve X0 = AX and find ONE particular solution Y0 (maybe by trial-and-error or other DE techniques)
to system Y0 = AY + b. Then you may just add this Y0 to the solution set of X0 = AX to get the general
solution of Y0 = AY + b.
2
```