Higher Engineering Mathematics

HYPERBOLIC FUNCTIONS 47

A

5.4 Solving equations involving

hyperbolic functions

Equations of the formachx+bshx=c, wherea,
bandcare constants may be solved either by:

(a) plotting graphs ofy=achx+bshxandy=c
and noting the points of intersection, or more
accurately,
(b) by adopting the following procedure:

(i) Change shxto

(

ex−e−x

2

)

and chxto

(

ex+e−x

2

)

(ii) Rearrange the equation into the form

pex+qe−x+r=0, wherep,qandrare

constants.

(iii) Multiply each term by ex, which produces

an equation of the formp(ex)^2 +rex+q= 0

(since (e−x)(ex)=e^0 =1)

(iv) Solve the quadratic equation

p(ex)^2 +rex+q=0 for exby factorising

or by using the quadratic formula.

(v) Given ex=a constant (obtained by solv-

ing the equation in (iv)), take Napierian

logarithms of both sides to give

x=ln (constant)

This procedure is demonstrated in Problems 12 to
14 following.

Problem 12. Solve the equation shx=3, cor-

rect to 4 significant figures.

Following the above procedure:

(i) shx=

(

ex−e−x

2

)

= 3

(ii) ex−e−x=6, i.e. ex−e−x− 6 = 0

(iii) (ex)^2 −(e−x)(ex)−6ex=0,

i.e. (ex)^2 −6ex− 1 = 0

(iv) ex=

−(−6)±

√

[(−6)^2 −4(1)(−1)]

2(1)

=

6 ±

√

40

2

=

6 ± 6. 3246

2

Hence ex= 6 .1623 or−0.1623

(v)x=ln 6.1623 orx=ln(−0.1623) which has

no solution since it is not possible in real

terms to find the logarithm of a negative num-

ber. Hencex=ln 6. 1623 =1.818, correct to 4

significant figures.

Problem 13. Solve the equation

2 .6chx+ 5 .1shx= 8 .73,

correct to 4 decimal places.

Following the above procedure:

(i) 2.6chx+ 5 .1shx= 8. 73

i.e. 2. 6

(

ex+e−x

2

)

+ 5. 1

(

ex−e−x

2

)

= 8. 73

(ii) 1.3ex+ 1 .3e−x+ 2 .55ex− 2 .55e−x= 8. 73

i.e. 3.85ex− 1 .25e−x− 8. 73 = 0

(iii) 3.85(ex)^2 − 8 .73ex− 1. 25 = 0

(iv) ex

=

−(− 8 .73)±

√

[(− 8 .73)^2 −4(3.85)(− 1 .25)]

2(3.85)

=

8. 73 ±

√

95. 463

7. 70

=

8. 73 ± 9. 7705

7. 70

Hence ex= 2 .4027 or ex=− 0. 1351

(v)x=ln 2.4027 orx=ln(−0.1351) which has no

real solution.

Hencex=0.8766, correct to 4 decimal places.

Problem 14. A chain hangs in the form given

byy=40 ch

x

40

. Determine, correct to 4 signifi-
cant figures, (a) the value ofywhenxis 25 and
(b) the value ofxwheny= 54 .30.

(a) y=40 ch

x

40

, and whenx=25,

y=40 ch

25

40

=40 ch 0. 625

= 40

(

e^0.^625 +e−^0.^625

2

)

=20(1. 8682 + 0 .5353)=48.07