Understanding Engineering Mathematics

Answer

(i) y=

x

2

+

C

x

(ii) y=x^3 +Cx^2 − 1

15.5 Second order linear homogeneous differential equations

If you found the previous section difficult then you may be approaching this section with
some trepidation – surely second order will be worse than first order? Relax. We only
consider a simple, but extremely important, type of second order equation and it turns out
that this is relatively straightforward to solve. The particular type of linear second order
equation that we will consider is one of the form

a

d^2 y

dx^2

+b

dy

dx

+cy=f(x) ( 15. 4 )

wherea,b,care constants (anda =0, otherwise it wouldn’t be second order!). For such
equations the basic tools that you need are simply solution of quadratics (including complex
roots) and solution of simultaneous linear equations. Equation (15.4) occurs everywhere in
science and engineering, most notably in the modelling of vibrating springs in a resisting
medium, and in electrical circuits.
We need some terminology. The equation:

a

d^2 y

dx^2

+b

dy

dx

+cy= 0 ( 15. 5 )

is the associatedhomogeneous equation, while equation (15.4), withf(x) =0, is the
inhomogeneousform. Thegeneral solutionto thehomogeneous equationis called the
complementary function, whileanysolution to equation (15.4) is called aparticular
integral.Thegeneral solutionof equation (15.4) isthe sum of the complementary
function and a particular integral.
For example the differential equation

d^2 y

dx^2

+ 3

dy

dx

+ 2 y= 2 ex

isinhomogeneous, and the correspondinghomogeneous formis

d^2 y

dx^2

+ 3

dy

dx

+ 2 y= 0

The general solution of this homogeneous equation is thecomplementary function of
the first, inhomogeneous, equation. In this case it happens to be (see below)

yc=Ae−x+Be−^2 x

whereAandBare two arbitrary constants.