Understanding Engineering Mathematics

(b) If the complementary function contains termsekx,xekx,...,

xmekx,takeyp(x)to be of the form

yp=Lxm+^1 ekx

(4) f(x)of the formekxcosωxorekxsinωx

(a) If these terms are not contained in the complementary function,

then simply takeyp(x)to be of the form

yp=Lekxcosωx+Mekxsinωx

(b) If terms of the form xrekxcosωx or xrekxsinωx,forr=

0 , 1 ,...,m, are contained in the complementary functions, then

include terms of the form

Lxm+^1 ekxcosωx+Mxm+^1 ekxsinωx

inyp.

Some obvious patterns may be seen in the above particular integrals. Essentially, if
the complementary function does not contain terms identical to the inhomogeneous term,
then the particular integral looks much the same as the inhomogeneous term. However, if
the inhomogeneous term occurs in the complementary function, then we must modify the
particular integral and this is done by multiplying by an appropriate power ofx.
In general, by superposition of terms we see that we can treat particular integrals for
expressions of the form

f(x)=P(x)ekxcosωx+Q(x)ekxsinωx

whereP(x),Q(x)are polynomials inx. This essentially comes about because any order
derivative of such an expression results in an expression of the same form.
Having found the particular integral, we can find the general solution by adding it to
the complementary function. Note again that when applying boundary or initial conditions
these must always be applied to the full solution – complementary function plus particular
integral – and not simply to the complementary function alone.

Exercises on 15.6

1. Find the solutions to each of the following second order equations, with the specified
   conditions.Remember to apply the conditions to the full solution – CFYPI.

(i) y′′+ 4 y′+ 3 y= 2 ex y( 0 )= 0 y′( 0 )= 1

(ii) y′′+ 4 y=x+ 1 y( 0 )= 0 y

(π

4

)

=^14

(iii) y′′+y=sin 2xy( 0 )= 0 y′( 0 )= 0

2. Solve the initial value problem

y′′− 4 y′+ 3 y= 3 xy( 0 )= 0 y′( 0 )= 0