Understanding Engineering Mathematics

Any particular solution of theinhomogeneousequation is aparticular integralof the
equation. By substituting it in the equation you should check that a particular integral in
this case is

yp=^13 ex

Thegeneral solution of the inhomogeneous equationis thus

y=yc+yp=Ae−x+Be−^2 x+^13 ex

We therefore have two jobs – to find the complementary function and to find a particular
integral. We first concentrate on finding the complementary function and look for general
solutions of equations of the form of equation (15.5). There is an additional result which
helpsushere:
Ify 1 andy 2 are two solutions of the homogeneous linear equation

a

d^2 y

dx^2

+b

dy

dx

+cy= 0

theny 1 +y 2 is also a solution. Thus, the sum (or any other linear combination) of two
solutions is also a solution. To see this important result in its most general form lety 1 ,y 2 be
two solutions of the DE and consider thelinear combinationy=αy 1 +βy 2. Substituting
this in the DE gives

a

d^2 y

dx^2

+b

dy

dx

+cy=a

d^2 (αy 1 +βy 2 )

dx^2

+b

d(αy 1 +βy 2 )

dx

+c(αy 1 +βy 2 )

=α

(

a

d^2 y 1

dx^2

+b

dy 1

dx

+cy 1

)

+β

(

a

d^2 y 2

dx^2

+b

dy 2

dx

+cy 2

)

=α( 0 )+β( 0 )= 0

So the general linear combinationy=αy 1 +βy 2 is also a solution.
In fact, finding the complementary function, i.e. the general solution to equation (15.5),
is not as difficult as might be thought. Note that if we puta=0 then we are back to the
simple equation

b

dy

dx

+cy= 0

and we know that this has an exponential solution (447
➤
). This encourages us to try a
similar exponential function for the second order equation.
We therefore take a trial solution

y=eλx

whereλis some constant parameter to be determined.
Substituting into the equation:

y′=λeλx,y′′=λ^2 eλx

gives
(aλ^2 +bλ+c)eλx= 0