The Chemistry Maths Book, Second Edition

11.2 Solution of a differential equation 317

EXAMPLE 11.2

(i) Show thaty 1 = 1 e

−x

1 + 11 is a solution of the first-order differential equation

We have. Therefore

(ii) Find the general solution of the second-order differential equation

.

We need to integrate twice:

0 Exercises 1–8

EXAMPLE 11.3A body falling under the influence of gravity

Let hbe the height of the body at time t. The (downward) force acting on the body is

F 1 = 1 −mgand, by Newton’s second law,F 1 = 1 md

2

h 2 dt

2

. The acceleration experienced

by the body is therefore

The first integration gives the velocity

and the second integration gives the height at time t

A particular solution in this case is given by the initial conditions for the system, the

heighth

0

and the velocityv

0

at timet 1 = 10. Thena 1 = 1 v

0

,b 1 = 1 h

0

, and

v(t) 1 = 1 −gt 1 + 1 v

0

,

0 Exercise 9

ht()=− gt t h+ +

1

2

2

00

v

h t()=− gt at b+ +

1

2

2

v()t

dh

dt

==−+gt a

dh

dt

g

2

2

=−

y

dy

dx

==−+dx x a dx x ax

















ZZ =− +

1

2

2

1

4

cos sin 2 ++b

dy

dx

dy

dx

== =−+ZZdx x dx x a

2

2

2

1

2

sin cos 2

dy

dx

x

2

2

=sin 2

dy

dx

yee

xx

+−=−













++













−=

−−

1110

dy

dx

e

x

=−

−

dy

dx

+−=y 10.