Understanding Engineering Mathematics

Dividing byk:

dn

kdt

=

dn

d(kt)

=n

and if we now introduce new variables:

x=kt,y=n

it becomes

dy

dx

=y( 15. 2 )

This is a good example of mathematical modelling. By changing the mathematical vari-
ables we have tidied up the equation and reduced the problem to its simplest mathematical
form. We always do this when we can.
The differential equation (15.2) is about as simple as you can get for a non-trivial
example. Yet it is an extremely important and basic equation. It occurs throughout the
theory of DEs, and it can be used directly to solve many more complicated equations,
including some of higher order (see Section 15.5).

Problem 15.1

Solve the equation (15.2) – i.e. findyas a function ofx.

The simplest approach is to turn the derivative upside down (235

➤

):

dx

dy

=

1

y

Now it is simply integrationwith respect toy

x=

∫

dx

dy

dy=

∫

1

y

dy=lny+C

whereCis an arbitrary constant.

So
lny=x−C
y=ex−C=e−Cex=Aex

whereAis another (positive) arbitrary constant.

So, a solution to

dy

dx

=yisy=AexwithAan arbitary constant.

Points to note are:

• The DE is said to befirst orderbecause the highest derivative involved
  is first order.


• The solution contains one arbitrary constant,A, giving us an infinity of
  solutions.


• Just one arbitrary constant arose precisely because to solve a first order
  DE we need to integrate just once.