Mathematical Methods for Physics and Engineering : A Comprehensive Guide

15.5 HINTS AND ANSWERS

15.36 Find the form of the solutions of the equation

dy

dx

d^3 y

dx^3

− 2

(

d^2 y

dx^2

) 2

+

(

dy

dx

) 2

=0

that havey(0) =∞.

[ You will need the result

∫z
cosechudu=−ln(cosechz+cothz). ]
15.37 Consider the equation

xpy′′+

n+3− 2 p

n− 1

xp−^1 y′+

(

p− 2

n− 1

) 2

xp−^2 y=yn,

in whichp=2andn>−1 butn= 1. For the boundary conditionsy(1) = 0 and

y′(1) =λ, show that the solution isy(x)=v(x)x(p−2)/(n−1),wherev(x)isgivenby

∫v(x)

0

dz

[

λ^2 +2zn+1/(n+1)

] 1 / 2 =lnx.

15.5 Hints and answers

15.1 The function isa(ω^20 −ω^2 )−^1 (cosωt−cosω 0 t); for moderatet,x(t) is a sine wave
of linearly increasing amplitude (tsinω 0 t)/(2ω 0 ); for largetit shows beats of
maximum amplitude 2(ω 02 −ω^2 )−^1.
15.3 Ignore the termy′^2 , compared with 1, in the expression forρ.y=0atx=0.
From symmetry,dy/dx=0atx=L/2.
15.5 General solutionf(t)=Ae−^6 t+Be−^2 t− 3 e−^4 t. (a) No solution, inconsistent
boundary conditions; (b)f(t)=2e−^6 t+e−^2 t− 3 e−^4 t.
15.7 The auxiliary equation has repeated roots and the RHS is contained in the
complementary function. The solution isy(x)=(A+Bx)e−x+2x^2 e−x.y(2) = 5e−^2.
15.9 (a) The auxiliary equation has roots 2, 2,−4; (A+Bx)exp2x+Cexp(− 4 x)+2x+1;
(b) multiply through by sinh 2∫ axand note that
cosech 2ax dx=(2a)−^1 ln(|tanhax|);y=B(sinh 2ax)^1 /^2 (|tanhax|)A.
15.11 Use Laplace transforms; writes(s+n)−^4 as (s+n)−^3 −n(s+n)−^4.
15.13 L[C(t)]=x 0 (s+8)/[s(s+2)(s+ 4)], yielding
C(t)=x 0 [1 +^12 exp(− 4 t)−^32 exp(− 2 t)].
15.15 The characteristic equation isλ^2 −λ−1=0.
un=[(1+

√

5)n−(1−

√

5)n]/(2n

√

5).

15.17 Fromu 4 andu 5 ,P=5,Q=−4.un=3/ 2 −5(−1)n/6+(−2)n/4+2n/ 12.
15.19 The general solution isA+B 2 n/^2 exp(i 3 πn/4)+C 2 n/^2 exp(i 5 πn/4). The initial values
imply thatA=1/ 5 ,B=(

√

5 /10) exp[i(π−φ)] andC=(

√

5 /10) exp[i(π+φ)].
15.21 This is Euler’s equation; settingx=exptproducesd^2 z/dt^2 − 2 dz/dt+z=expt,
with complementary function (A+Bt)exptand particular integralt^2 (expt)/2;
y(x)=x+[xlnx(1 + lnx)]/2.
15.23 After multiplication through byx^2 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 /x^2.
15.25 Given the boundary conditions, it is better to work with sinhxand sinh(1−x)
than withe±x;G(x, ξ)=−[sinh(1−ξ)sinhx]/sinh 1 forx<ξand−[sinh(1−
x)sinhξ]/sinh 1 forx>ξ.
15.27 Follow the method of subsection 15.2.5, but using general rather than specific
functions.
15.29 G(t, τ)=0fort<τandκ−^1 e−(t−τ)sin[κ(t−τ)] fort>τ. For a unit step input,
x(t)=(1+κ^2 )−^1 (1−e−tcosκt−κ−^1 e−tsinκt). Both transforms are equivalent to
s[(s+1)^2 +κ^2 )] ̄x=1.