Math 2C Noonburg - Chapter 2 Notes 2/14/15
1. (2.3#15) In §1.3, the velocity v(t) of a free-falling body was shown to satisfy the differential equation
mv 0 (t) = mg − k(v(t)p
where m is the mass of the body in kilograms, g is the gravitational acceleration and p is a real constant. If we
assume p = 1 the sage can be used to produce and exact solution.
(a) For what value(s) of p is the equation linear in v(t)?
ANS: p = 0 or p = 1.
k
(b) Solve the equation with p = 1,
= 0.6, and g = 9.8m/ sec2 .
m
ANS: First convert to standard form:
v 0 + 0.6v = 9.8
d 0.6t This is then first-order, linear so we can multiply through by µ = e0.6t so the equation becomes
e v =
dt
49
49
9.8e0.6t ⇔ e0.6t v = e0.6t + C ⇔ v =
+ Ce−0.6t
3
3
(c) What is the value of the terminal velocity; that is, lim v(t)? Convert the velocity from meters per second
t→∞
to miles per hour.
1miles · 3600 sec
49
. To convert from meters/sec to miles/hour, multiply by
to get
ANS: As t → ∞, v →
3
1609metershour
≈ 36.5 mph.
(d) Does the terminal velocity depend on the initial velocity v(0).
ANS: Evidently it does not.
2. (2.3#16) In Exercises 1.3 (under ENGINEERING), the RL circuit problem
i0 (t) + 3i(t) = 10 sin(t)
was given and you were asked where to find the value of i(0) that makes the solution be i(t) = 3 sin(t) − cos(t)).
Use the method described in this section to solve the equation,
and determine
the correct value of i(0). (The
R at
a
sin(bt)
−
b
cos(bt)
following integral formula may help: e sin(bt)dt = eat
.) Use your CAS to draw a slope
a2 + b2
filed for the equation and plot the solution using the initial condition you found.
d st
ANS: The integrating factor is µ = e3t and leads to
e i(t) = 10 sin(t)e3t . In 1B you learned how to do this
dt
by integrating by parts twice. I’ve learned a new way by watching MIT Video, 6:30. Use the euler formula, eIt =
cos(t) + I sin(t), (oh dear, that’s the imaginary I, not the current...) and think of 10 sin(t)e3t = Im 10eIt e3t =
Im 10e(3+I)t , that is, the imaginary part of 10e3t
(cos(t) + I sin(t)).
Then integrating is relatively easy (no IbyP
R
10
e(3+I)t + C = Im (3 − I)e3t (cos(t) + I sin(t)) + C,
is necessary): est i(t) = Im 10e(3+I)t dt = Im
3+I
10
3−I
10(3 − I)
where we’ve used the conversion
·
=
= 3 − I. Now multiply the two complex num3+I 3−I
9+1
bers and, voil, the imaginary part is our integral: est i(t) = Im e3t (3 cos(t) + sin(t) + I(3 sin(t) − cos(t)) + C =
e3t (3 sin(t) − cos(t)) + C, which comports with Noonburg’s formula and yields i(t) = 3 sin(t) − cos(t) + Ce−3t
And now for a little side bar on proving the identity:
A cos(θ) + B sin(θ) =
p
A2 + B 2 cos(θ − φ)
where φ = arctan(B/A). There are three different proofs for this, but for simply remembering it, this picture is
useful:
Math 2C
Noonburg - Chapter 2 Notes - Page 2 of 3
2/14/15
In high school trigonometry we prove this by using the addition identity to expand the right side and then
equating coefficients arrive at an aha. But that requires knowing the right hand side a priori, which is kind of
cheating.
Another way to see this is to start with the left side and view as the dot product of the unit vector ûθ = hcos θ, sin θi
with hA, Bi, which is illustrated in this diagram:
Using the formula for the dot product we have that hcos θ, sin θi · hA, Bi = |hA, Bi| cos(θ − φ), pretty sweet, huh?
A third way is to use complex numbers and view the left hand side as the real part of the product (A−Bi)(cos θ +
i sin θ). The minus sign in the
is needed because i2 = −1. Now convert these factors to polar repre√ first factor
−iφ
2
2
sentations, where A − Bi = A + B e
and the
simply √
eiθ . Multiplying these together with the law
√ second is−iφ
iθ
of exponents we have (A − Bi)(cos θ + i sin θ) = A2 + B 2 e e = A2 + B 2 ei(θ−φ) whose real part is what we
want.
ok, so back to the question at hand. We want i(t) = 3 sin(t) − cos(t)), which is what you get if you choose C = 0.
Didn’t we already solve that once? Anyway, here is a way to plot this with the slope field in Sage:
t,i=var(’t i’)
f(t,i)=10*sin(t)-3*i
p0=plot_slope_field(f(t,i),(t, 0, 15), (i, -5, 5))
p1=plot(3*sin(t)-cos(t),(0,15))
show(p0+p1)
Math 2C
Noonburg - Chapter 2 Notes - Page 3 of 3
2/14/15
3. (2.4#3) Does the equation x0 = x2 − t have a unique solution through every initial point (t0 , x0 )? Can solution
curves ever intersect for this differential equation? Is so, where?
ANS: The function f (t, x) = x2 − t is continuous everywhere and its first partial derivative with respect to x,
δf
= 2x is also continuous everywhere. Therefore, by the existence and uniqueness theorem, solutions for any
δx
initial value will be unique and non-intersecting.
4. Does Theorem 2.1 imply that the solution of the IVP x0 = x2 − t, x(0) = 1.0 is defined for all t? What can you
say bout the size of the interval for which it exists?
No, just because we are guaranteed a solution for all initial values, does not mean that the domain of each solution
will be −∞ < t < ∞.