Exercises

15.4 EXERCISES
15.3.6 Equations having y = Aex as a solution
Finally, we note that if any general (linear or non-linear) nth-order ODE is
satisfied identically by assuming that
dn y
dy
= ··· = n
(15.88)
dx
dx
then y = Aex is a solution of that equation. This must be so because y = Aex is
a non-zero function that satisfies (15.88).
y=
Find a solution of
(x2 + x)
dy d2 y
dy
− x2 y
−x
dx dx2
dx
dy
dx
2
= 0.
(15.89)
Setting y = dy/dx = d2 y/dx2 in (15.89), we obtain
(x2 + x)y 2 − x2 y 2 − xy 2 = 0,
which is satisfied identically. Therefore y = Aex is a solution of (15.89); this is easily
verified by directly substituting y = Aex into (15.89). Solution method. If the equation is satisfied identically by making the substitutions
y = dy/dx = · · · = dn y/dxn then y = Aex is a solution.
15.4 Exercises
15.1
15.2
A simple harmonic oscillator, of mass m and natural frequency ω0 , experiences
an oscillating driving force f(t) = ma cos ωt. Therefore, its equation of motion is
d2 x
+ ω02 x = a cos ωt,
dt2
where x is its position. Given that at t = 0 we have x = dx/dt = 0, find the
function x(t). Describe the solution if ω is approximately, but not exactly, equal
to ω0 .
Find the roots of the auxiliary equation for the following. Hence solve them for
the boundary conditions stated.
df
d2 f
+2
+ 5f = 0, with f(0) = 1, f (0) = 0.
dt2
dt
d2 f
df
(b) 2 + 2
+ 5f = e−t cos 3t, with f(0) = 0, f (0) = 0.
dt
dt
The theory of bent beams shows that at any point in the beam the ‘bending
moment’ is given by K/ρ, where K is a constant (that depends upon the beam
material and cross-sectional shape) and ρ is the radius of curvature at that point.
Consider a light beam of length L whose ends, x = 0 and x = L, are supported
at the same vertical height and which has a weight W suspended from its centre.
Verify that at any point x (0 ≤ x ≤ L/2 for definiteness) the net magnitude of
the bending moment (bending moment = force × perpendicular distance) due to
the weight and support reactions, evaluated on either side of x, is Wx/2.
(a)
15.3
523
HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS
If the beam is only slightly bent, so that (dy/dx)2 1, where y = y(x) is the
downward displacement of the beam at x, show that the beam profile satisfies
the approximate equation
15.4
d2 y
Wx
=−
.
dx2
2K
By integrating this equation twice and using physically imposed conditions on
your solution at x = 0 and x = L/2, show that the downward displacement at
the centre of the beam is W L3 /(48K).
Solve the differential equation
df
d2 f
+6
+ 9f = e−t ,
dt2
dt
15.5
subject to the conditions f = 0 and df/dt = λ at t = 0.
Find the equation satisfied by the positions of the turning points of f(t) and
hence, by drawing suitable sketch graphs, determine the number of turning points
the solution has in the range t > 0 if (a) λ = 1/4, and (b) λ = −1/4.
The function f(t) satisfies the differential equation
df
d2 f
+8
+ 12f = 12e−4t .
dt2
dt
For the following sets of boundary conditions determine whether it has solutions,
and, if so, find them:
√
2) = 0;
(a) f(0) = 0, f (0) = 0, f(ln √
(b) f(0) = 0, f (0) = −2, f(ln 2) = 0.
15.6
Determine the values of α and β for which the following four functions are
linearly dependent:
y1 (x) = x cosh x + sinh x,
y2 (x) = x sinh x + cosh x,
y3 (x) = (x + α)ex ,
y4 (x) = (x + β)e−x .
15.7
You will find it convenient to work with those linear combinations of the yi (x)
that can be written the most compactly.
A solution of the differential equation
dy
d2 y
+2
+ y = 4e−x
dx2
dx
15.8
takes the value 1 when x = 0 and the value e−1 when x = 1. What is its value
when x = 2?
The two functions x(t) and y(t) satisfy the simultaneous equations
dx
− 2y = − sin t,
dt
dy
+ 2x = 5 cos t.
dt
Find explicit expressions for x(t) and y(t), given that x(0) = 3 and y(0) = 2.
Sketch the solution trajectory in the xy-plane for 0 ≤ t < 2π, showing that
the trajectory crosses itself at (0, 1/2) and passes through the points (0, −3) and
(0, −1) in the negative x-direction.
524
15.4 EXERCISES
15.9
Find the general solutions of
(a)
(b)
15.10
dy
d3 y
− 12
+ 16y = 32x − 8,
dx3
dx
1 dy
d 1 dy
+ (2a coth 2ax)
= 2a2 ,
dx y dx
y dx
where a is a constant.
Use the method of Laplace transforms to solve
df
d2 f
+ 6f = 0,
f(0) = 1, f (0) = −4,
+5
dt2
dt
2
df
df
(b)
+2
+ 5f = 0,
f(0) = 1, f (0) = 0.
dt2
dt
The quantities x(t), y(t) satisfy the simultaneous equations
(a)
15.11
ẍ + 2nẋ + n2 x = 0,
ÿ + 2nẏ + n2 y = µẋ,
where x(0) = y(0) = ẏ(0) = 0 and ẋ(0) = λ. Show that
y(t) = 12 µλt2 1 − 13 nt exp(−nt).
15.12
Use Laplace transforms to solve, for t ≥ 0, the differential equations
ẍ + 2x + y = cos t,
ÿ + 2x + 3y = 2 cos t,
15.13
15.14
which describe a coupled system that starts from rest at the equilibrium position.
Show that the subsequent motion takes place along a straight line in the xy-plane.
Verify that the frequency at which the system is driven is equal to one of the
resonance frequencies of the system; explain why there is no resonant behaviour
in the solution you have obtained.
Two unstable isotopes A and B and a stable isotope C have the following decay
rates per atom present: A → B, 3 s−1 ; A → C, 1 s−1 ; B → C, 2 s−1 . Initially a
quantity x0 of A is present, but there are no atoms of the other two types. Using
Laplace transforms, find the amount of C present at a later time t.
For a lightly damped (γ < ω0 ) harmonic oscillator driven at its undamped
resonance frequency ω0 , the displacement x(t) at time t satisfies the equation
dx
d2 x
+ 2γ
+ ω02 x = F sin ω0 t.
dt2
dt
Use Laplace transforms to find the displacement at a general time if the oscillator
starts from rest at its equilibrium position.
(a) Show that ultimately the oscillation has amplitude F/(2ω0 γ), with a phase
lag of π/2 relative to the driving force per unit mass F.
(b) By differentiating the original equation, conclude that if x(t) is expanded as
a power series in t for small t, then the first non-vanishing term is Fω0 t3 /6.
Confirm this conclusion by expanding your explicit solution.
15.15
The ‘golden mean’, which is said to describe the most aesthetically pleasing
proportions for the sides of a rectangle (e.g. the ideal picture frame), is given
by the limiting value of the ratio of successive terms of the Fibonacci series un ,
which is generated by
un+2 = un+1 + un ,
with u0 = 0 and u1 = 1. Find an expression for the general term of the series and
525
HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS
15.16
15.17
verify that the golden mean is equal to the larger root of the recurrence relation’s
characteristic equation.
In a particular scheme for numerically modelling one-dimensional fluid flow, the
successive values, un , of the solution are connected for n ≥ 1 by the difference
equation
c(un+1 − un−1 ) = d(un+1 − 2un + un−1 ),
where c and d are positive constants. The boundary conditions are u0 = 0 and
uM = 1. Find the solution to the equation, and show that successive values of un
will have alternating signs if c > d.
The first few terms of a series un , starting with u0 , are 1, 2, 2, 1, 6, −3. The series
is generated by a recurrence relation of the form
un = P un−2 + Qun−4 ,
where P and Q are constants. Find an expression for the general term of the
series and show that, in fact, the series consists of two interleaved series given by
u2m =
u2m+1 =
15.18
2
3
7
3
+ 13 4m ,
− 13 4m ,
for m = 0, 1, 2, . . . .
Find an explicit expression for the un satisfying
un+1 + 5un + 6un−1 = 2n ,
15.19
given that u0 = u1 = 1. Deduce that 2n − 26(−3)n is divisible by 5 for all
non-negative integers n.
Find the general expression for the un satisfying
un+1 = 2un−2 − un
with u0 = u1 = 0 and u2 = 1, and show that they can be written in the form
3πn
1 2n/2
un = − √ cos
−φ ,
5
4
5
15.20
where tan φ = 2.
Consider the seventh-order recurrence relation
un+7 − un+6 − un+5 + un+4 − un+3 + un+2 + un+1 − un = 0.
Find the most general form of its solution, and show that:
(a) if only the four initial values u0 = 0, u1 = 2, u2 = 6 and u3 = 12, are specified,
then the relation has one solution that cycles repeatedly through this set of
four numbers;
(b) but if, in addition, it is required that u4 = 20, u5 = 30 and u6 = 42 then the
solution is unique, with un = n(n + 1).
15.21
Find the general solution of
15.22
dy
d2 y
−x
+ y = x,
dx2
dx
given that y(1) = 1 and y(e) = 2e.
Find the general solution of
x2
(x + 1)2
dy
d2 y
+ 3(x + 1)
+ y = x2 .
dx2
dx
526
15.4 EXERCISES
15.23
Prove that the general solution of
(x − 2)
is given by
y(x) =
15.24
d2 y
dy
4y
+3
+ 2 =0
dx2
dx
x
1
2
1
+ cx2 .
k
−
2
(x − 2)
3x 2
Use the method of variation of parameters to find the general solutions of
d2 y
d2 y
dy
− y = xn , (b)
−2
+ y = 2xex .
dx2
dx2
dx
Use the intermediate result of exercise 15.24(a) to find the Green’s function that
satisfies
(a)
15.25
d2 G(x, ξ)
− G(x, ξ) = δ(x − ξ)
dx2
15.26
with
G(0, ξ) = G(1, ξ) = 0.
Consider the equation
F(x, y) = x(x + 1)
d2 y
dy
+ (2 − x2 )
− (2 + x)y = 0.
dx2
dx
(a) Given that y1 (x) = 1/x is one of its solutions, find a second linearly independent one,
(i) by setting y2 (x) = y1 (x)u(x), and
(ii) by noting the sum of the coefficients in the equation.
(b) Hence, using the variation of parameters method, find the general solution
of
F(x, y) = (x + 1)2 .
15.27
Show generally that if y1 (x) and y2 (x) are linearly independent solutions of
dy
d2 y
+ p(x)
+ q(x)y = 0,
dx2
dx
with y1 (0) = 0 and y2 (1) = 0, then the Green’s function G(x, ξ) for the interval
0 ≤ x, ξ ≤ 1 and with G(0, ξ) = G(1, ξ) = 0 can be written in the form
#
y1 (x)y2 (ξ)/W (ξ) 0 < x < ξ,
G(x, ξ) =
y2 (x)y1 (ξ)/W (ξ) ξ < x < 1,
15.28
where W (x) = W [y1 (x), y2 (x)] is the Wronskian of y1 (x) and y2 (x).
Use the result of the previous exercise to find the Green’s function G(x, ξ) that
satisfies
dG
d2 G
+3
+ 2G = δ(x − x),
dx2
dx
in the interval 0 ≤ x, ξ ≤ 1, with G(0, ξ) = G(1, ξ) = 0. Hence obtain integral
expressions for the solution of
#
dy
d2 y
0 0 < x < x0 ,
+
3
+
2y
=
1 x0 < x < 1,
dx2
dx
distinguishing between the cases (a) x < x0 , and (b) x > x0 .
527
HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS
15.29
The equation of motion for a driven damped harmonic oscillator can be written
ẍ + 2ẋ + (1 + κ2 )x = f(t),
15.30
with κ = 0. If it starts from rest with x(0) = 0 and ẋ(0) = 0, find the corresponding
Green’s function G(t, τ) and verify that it can be written as a function of t − τ
only. Find the explicit solution when the driving force is the unit step function,
i.e. f(t) = H(t). Confirm your solution by taking the Laplace transforms of both
it and the original equation.
Show that the Green’s function for the equation
d2 y
y
+ = f(x),
dx2
4
subject to the boundary conditions y(0) = y(π) = 0, is given by
#
−2 cos 12 x sin 12 z 0 ≤ z ≤ x,
G(x, z) =
−2 sin 12 x cos 12 z x ≤ z ≤ π.
15.31
Find the Green’s function x = G(t, t0 ) that solves
dx
d2 x
+α
= δ(t − t0 )
dt2
dt
under the initial conditions x = dx/dt = 0 at t = 0. Hence solve
dx
d2 x
+α
= f(t),
dt2
dt
15.32
where f(t) = 0 for t < 0.
Evaluate your answer explicitly for f(t) = Ae−at (t > 0).
Consider the equation
d2 y
+ f(y) = 0,
dx2
where f(y) can be any function.
(a) By multiplying through by dy/dx, obtain the general solution relating x and
y.
(b) A mass m, initially at rest at the point x = 0, is accelerated by a force
x
.
f(x) = A(x0 − x) 1 + 2 ln 1 −
x0
Its equation of motion is m d2 x/dt2 = f(x). Find x as a function of time, and
show that ultimately the particle has travelled a distance x0 .
15.33
Solve
2y
15.34
2
d3 y
dy d2 y
dy
+2 y+3
+2
= sin x.
3
2
dx
dx dx
dx
Find the general solution of the equation
x
15.35
d3 y
d2 y
+ 2 2 = Ax.
dx3
dx
Express the equation
dy
d2 y
2
+ 4x
+ (4x2 + 6)y = e−x sin 2x
dx2
dx
in canonical form and hence find its general solution.
528