Mathematical Methods for Physics and Engineering : A Comprehensive Guide

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS

Solve

d^2 y

dx^2

− 3

dy

dx

+2y=2e−x, (15.33)

subject to the boundary conditionsy(0) = 2,y′(0) = 1.

Taking the Laplace transform of (15.33) and using the table of standard results we obtain

s^2 ̄y(s)−sy(0)−y′(0)− 3 [s ̄y(s)−y(0)]+2 ̄y(s)=

2

s+1

,

which reduces to

(s^2 − 3 s+2) ̄y(s)− 2 s+5=

2

s+1

. (15.34)

Solving this algebraic equation for ̄y(s), the Laplace transform of the required solution to
(15.33), we obtain

̄y(s)=

2 s^2 − 3 s− 3

(s+1)(s−1)(s−2)

=

1

3(s+1)

+

2

s− 1

−

1

3(s−2)

, (15.35)

where in the final step we have used partial fractions. Taking the inverse Laplace transform
of (15.35), again using table 13.1, we find the specific solution to (15.33) to be

y(x)=^13 e−x+2ex−^13 e^2 x.

Note that if the boundary conditions in a problem are given as symbols, rather

than just numbers, then the step involving partial fractions can often involve

a considerable amount of algebra. The Laplace transform method is also very

convenient for solving sets ofsimultaneouslinear ODEs with constant coefficients.

Two electrical circuits, both of negligible resistance, each consist of a coil having self-

inductanceLand a capacitor having capacitanceC. The mutual inductance of the two

circuits isM. There is no source of e.m.f. in either circuit. Initially the second capacitor

is given a chargeCV 0 , the first capacitor being uncharged, and at timet=0aswitchin

the second circuit is closed to complete the circuit. Find the subsequent current in the first

circuit.

Subject to the initial conditionsq 1 (0) = ̇q 1 (0) = ̇q 2 (0) = 0 andq 2 (0) =CV 0 =V 0 /G,say,
we have to solve

Lq ̈ 1 +M ̈q 2 +Gq 1 =0,

M ̈q 1 +L ̈q 2 +Gq 2 =0.

On taking the Laplace transform of the above equations, we obtain

(Ls^2 +G) ̄q 1 +Ms^2 ̄q 2 =sMV 0 C,

Ms^2 q ̄ 1 +(Ls^2 +G) ̄q 2 =sLV 0 C.

Eliminating ̄q 2 and rewriting as an equation for ̄q 1 , we find

̄q 1 (s)=

MV 0 s

[(L+M)s^2 +G][(L−M)s^2 +G]

=

V 0

2 G

[

(L+M)s

(L+M)s^2 +G

−

(L−M)s

(L−M)s^2 +G

]

.