Download Part 1: This is part of the homework that is due tomorrow (Friday

Part 1: This is part of the homework that is due tomorrow (Friday October 28) at the
start of class.
Problem w9.1 in the homework tests whether you can find bases for the solution spaces
of homogeneous linear di↵erential equations with constant coefficients (section 5.3), via the
method that looks at the roots of characteristic polynomials. The one case we haven’t
covered in detail yet is when the characteristic polynomial has complex roots r = a ± bi.
In this case the two complex function solutions e(a+bi)x , e(a bi)x yield the two real function
solutions eax cos(bx), eax sin(bx).
1. (a) (this is w9.1a) Find the general solution y(x) to the di↵erential equation
y (3)
5y 00 + 3y 0 + 9y = 0
Hint: Find a real root r1 of the cubic characteristic polynomial then divide the
characteristic polynomial by r r1 to get the quotient quadratic, which will factor
easily.
(b) (this is w9.1b) Find the general solution x(t) to the di↵erential equation
x00 (t) + 4x0 (t) + 29x(t) = 0.
Hint: completing the square to get the roots of the characteristic polynomial works
well here - probably better than the quadratic formula.
(c) (this is w9.1d) Find the general solution y(x) to the di↵erential equation
y (4)
8y 0 = 0.
Math 2250 Lab 8
Due Date: Nov. 3
Name/Unid:
1. The figure below depicts a typical mass-spring-dashpot system; here, m denotes the
mass of the oscillating object (in kilograms), k denotes the spring constant (in Newtons/meter), and c denotes the damping constant. We use x(t) to denote the (horizontal) position of the mass at time t (with x = 0 being its equilibrium position); positive
directions of x imply the mass is to the right of its equilibrium position. Newton’s second
law yields the di↵erential equation
mx00 =
kx
cx0 + F (t),
(1)
where kx is the force exerted by the spring, proportional to displacement from equilibrium, cx0 is the linear drag force proportional to velocity, and F (t) is any external
force depending on time.
Figure 1:
We can rewrite this DE in linear form
mx00 + cx0 + kx = F (t).
(2)
In the case of no external forcing we get the homogeneous DE
mx00 + cx0 + kx = 0.
(3)
In the following problem we fix the values m = 4 and k = 36 and set F = 0. However,
the value of the damping constant c will vary. In this setup, x(t) satisfies the
ODE 4x00 (t) + cx0 (t) + 36x(t) = 0, which after dividing through by the mass 4 is equal
to
c
x00 (t) + · x0 (t) + 9 · x(t) = 0
4
In the di↵erent parts of this problem, “IVP” will refer to this ODE subject to the initial
conditions
x(0) = 2 and x0 (0) = 3
Page 2
(a) Solve the IVP when there is no damping, i.e., when c = 0. After finding your
solution in linear combination form, convert into amplitude-phase form. Identify
the numerical values for amplitude, phase angle, and time delay. (See Friday’s class
notes for a review of amplitude-phase form, or text section 5.4.)
Page 3
(b) Solve the IVP when c = 0.8 (we refer to the situation when c is relatively small
as under-damped ). Instructions: When finding the roots of the characteristic
polynomial round o↵ to five digits.
(c) A value of the damping constant induces so-called critical damping whenever the
associated characteristic polynomial has a double real root. Confirm that c = 24
leads to critical damping, and solve the IVP in this case.
(d) Finally, solve the IVP when c = 30 (here, we are in the situation of over-damping),
i.e the solution is a linear combination of time-decaying exponential functions.
(e) Use Maple or other software (Matlab, Wolfram alpha, etc) to create a display containing the graphs of all four solutions above, on the interval 0  t  3. Print out
a copy, and label which graph corresponds to which solution.
Page 4
2. Energy in a mass-spring-damper system: Let x(t) be the position of a mass m attached
to a spring with Hooke’s constant k and damping piston with constant c, yielding the
di↵erential equation
mx00 + cx0 + kx = f (t),
where f (t) is an external forcing on the mass. We wish to account for the total energy
of the mass-spring configuration, neglecting the heat energy loss due to damping. We
define the total energy E(t) to be the sum of kinetic and potential energy. Potential
energy P E(t) is stored by the compressed or stretched spring, and is the work done to
stretch/compress it as the mass moves from from equilibrium x = 0, to position x:
Z x
k
P E(t) =
kudu = x2
2
0
As usual, the kinetic energy of the mass is
KE(t) =
m 0 2
(x ) .
2
The sum is the total energy
E(t) = P E(t) + KE(t) =
⌘
1⇣ 2
kx + m(x0 )2
2
(a) Take the derivative of E(t) with respect to time. Use the chain rule on the right side
of the equation. Then simplify your result, so that you get a formula for E 0 (t) that
only depends on the forcing f (t), the velocity x0 (t), and the damping coefficient c.
Page 5
(b) Assume there is no external forcing, i.e. f = 0. In this case, what condition(s) on
m,c,k guarantee that the energy in the system is constant (i.e., dE/dt = 0)?
(c) Set f (t) = 0, m = 1, c = 2, and k = 5. Set initial conditions to be x(0) = 2 and
x0 (0) = 0. How long will it take for the system to loose 80% of its initial energy? To
solve, find the solution x(t) to the DE and use this solution to compute the energy
function E explicitly.
(d) Plot the energy curve E(t) and describe its behavior. Explain in words the mechanism that drives the energy picture that you observe, and why the energy is not
decreasing initially, but then decreases more rapidly later on.
Page 6
3. (Where the magic algorithms of Section 5.3 come from.) Let D represent di↵erentiation
d
d2
with respect to x, that is D = dx
, D2 = dx
Let a1 , a0 be scalars. Use I for
2 , etc.
the identity operator, I(y) = y. Then we can rewrite our familiar second order linear
operator L as a linear combination of the operators I, D, D2 (see page 341 of our text):
Ly = y 00 + a1 y 0 + a0 y
= D2 (y) + a1 D(y) + a0 I(y)
⇥
⇤
= D2 + a1 D + a0 I (y)
.
So we may rewrite the homogeneous DE
y 00 + a1 y 0 + a0 y = 0
as
⇥
⇤
D2 + a1 D + a0 I (y) = 0.
If the characteristic polynomial p(r) = r2 + a1 r + a0 factors as
p(r) = (r
a)(r
b)
then the operator L factors as a composition of linear first order operators (in either
order):
D2 + a1 D + a0 I = [D aI] [D bI] = [D bI] [D aI].
For distinct roots a 6= b the basis functions are y1 = eax and y2 = ebx because
[D
aI]eax = 0
and
[D
(a) Verify that
[D
aI]eax = 0.
Page 7
bI]ebx = 0.
(b) Now suppose instead that the characteristic polynomial has a double root a = b,
i.e. p(r) = (r a)2 . Then
L = D2 + a1 D + a0 I = [D
aI] [D
aI].
Show that
[D
aI]xeax = eax .
(c) Use part (a) and composition of operators to deduce that
[D
aI] [D
aI]xeax = 0
(This explains why for double roots our homogeneous DE solution space basis is
{eax , xeax }.)
(d) Show that for any di↵erentiable function f (x),
[D
aI]f (x)eax = f 0 (x)eax .
(e) Explain using (d) why if the operator L of degree at least three has a factor
[D
aI] [D
aI] [D
aI] = [D
aI]3
Then the three functions eax , xeax , x2 eax all solve L(y) = 0.
Page 8