Hints and answers

15.5 HINTS AND ANSWERS
15.36
15.37
Find the form of the solutions of the equation
2 2 2
dy d3 y
dy
dy
−
2
+
=0
dx dx3
dx2
dx
that have y(0) = ∞.
z
[ You will need the result
cosech u du = − ln(cosech z + coth z). ]
Consider the equation
2
n + 3 − 2p p−1 p−2
xp−2 y = y n ,
xp y +
x y +
n−1
n−1
in which p = 2 and n > −1 but n = 1. For the boundary conditions y(1) = 0 and
y (1) = λ, show that the solution is y(x) = v(x)x(p−2)/(n−1) , where v(x) is given by
v(x)
dz
1/2 = ln x.
0
λ2 + 2z n+1 /(n + 1)
15.5 Hints and answers
15.1
15.3
15.5
15.7
15.9
15.11
15.13
15.15
15.17
15.19
15.21
15.23
15.25
15.27
15.29
The function is a(ω02 − ω 2 )−1 (cos ωt − cos ω0 t); for moderate t, x(t) is a sine wave
of linearly increasing amplitude (t sin ω0 t)/(2ω0 ); for large t it shows beats of
maximum amplitude 2(ω02 − ω 2 )−1 .
Ignore the term y 2 , compared with 1, in the expression for ρ. y = 0 at x = 0.
From symmetry, dy/dx = 0 at x = L/2.
General solution f(t) = Ae−6t + Be−2t − 3e−4t . (a) No solution, inconsistent
boundary conditions; (b) f(t) = 2e−6t + e−2t − 3e−4t .
The auxiliary equation has repeated roots and the RHS is contained in the
complementary function. The solution is y(x) = (A+Bx)e−x +2x2 e−x . y(2) = 5e−2 .
(a) The auxiliary equation has roots 2, 2, −4; (A+Bx) exp 2x+C exp(−4x)+2x+1;
sinh 2ax and note that
(b) multiply through by
cosech 2ax dx = (2a)−1 ln(| tanh ax|); y = B(sinh 2ax)1/2 (| tanh ax|)A .
Use Laplace transforms; write s(s + n)−4 as (s + n)−3 − n(s + n)−4 .
L [C(t)] = x0 (s + 8)/[s(s + 2)(s + 4)], yielding
C(t) = x0 [1 + 12 exp(−4t) − 32 exp(−2t)].
2
The characteristic
equation
− λ − 1 = 0.
√
√ is λ √
un = [(1 + 5)n − (1 − 5)n ]/(2n 5).
From u4 and u5 , P = 5, Q = −4. un = 3/2 − 5(−1)n /6 + (−2)n /4 + 2n /12.
n/2
exp(i3πn/4)+C2n/2 exp(i5πn/4).
The initial values
The general solution is A+B2
√
√
imply that A = 1/5, B = ( 5/10) exp[i(π − φ)] and C = ( 5/10) exp[i(π + φ)].
This is Euler’s equation; setting x = exp t produces d2 z/dt2 − 2 dz/dt + z = exp t,
with complementary function (A + Bt) exp t and particular integral t2 (exp t)/2;
y(x) = x + [x ln x(1 + ln x)]/2.
After multiplication through by x2 the coefficients are such that this is an
exact equation. The resulting first-order equation, in standard form, needs an
integrating factor (x − 2)2 /x2 .
Given the boundary conditions, it is better to work with sinh x and sinh(1 − x)
than with e±x ; G(x, ξ) = −[sinh(1 − ξ) sinh x]/ sinh 1 for x < ξ and −[sinh(1 −
x) sinh ξ]/ sinh 1 for x > ξ.
Follow the method of subsection 15.2.5, but using general rather than specific
functions.
G(t, τ) = 0 for t < τ and κ−1 e−(t−τ) sin[κ(t − τ)] for t > τ. For a unit step input,
x(t) = (1 + κ2 )−1 (1 − e−t cos κt − κ−1 e−t sin κt). Both transforms are equivalent to
s[(s + 1)2 + κ2 )]x̄ = 1.
529
HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS
15.31
15.33
15.35
15.37
Use continuity and the step condition on ∂G/∂t at t = t0 to show that
G(t, t0 ) = α−1 {1 − exp[α(t0 − t)]} for 0 ≤ t0 ≤ t;
x(t) = A(α − a)−1 {a−1 [1 − exp(−at)] − α−1 [1 − exp(−αt)]}.
The LHS of the equation is exact for two stages of integration and then needs
an integrating factor exp x; 2y d2 y/dx2 + 2y dy/dx + 2(dy/dx)2 ; 2y dy/dx + y 2 =
d(y 2 )/dx + y 2 ; y 2 = A exp(−x) + Bx + C − (sin x − cos x)/2.
2
Follow the method of subsection 15.2.6; u(x) = e−x and v(x) satisfies v + 4v =
sin 2x, for which a particular integral is (−x cos 2x)/4. The general solution is
2
y(x) = [A sin 2x + (B − 14 x) cos 2x]e−x .
The equation is isobaric, with y of weight m, where m + p − 2 = mn; v(x)
satisfies x2 v + xv = v n . Set x = et and v(x) = u(t), leading to u = un with
u(0) = 0, u (0) = λ. Multiply both sides by u to make the equation exact.
530