Understanding Engineering Mathematics

We have

yp′′−yp′− 2 yp=L(x− 2 )e−x−L( 1 −x)e−x− 2 Lxe−x

=− 3 Le−x=the RHS=e−x

SoLxe−xis indeed a solution of the DE ifL=−^13 and therefore a particular solution is

yp=−^13 xe−x.
Now, you may feel uncomfortable with all this guesswork. Don’t worry, it can all be put
on a formal basis and derived rigorously, but we do not have the space to go into it here.
So, we will simply collect all the necessary results together and state them without proof,
for your reference. The method to be described actually works for arbitrary order linear
equations with constant coefficients and so we will formulate it in quite general terms.
To specialise to the second order case we have been considering so far just taken=2.
However, you may meet the general case of higher ordernin, for example, control systems.
So, consider the generalnth order linear differential equation with constant coefficients

(y(r)(x)denotes therth order derivative

dry

dxr

):

any(n)(x)+an− 1 y(n−^1 )(x)+···+a 1 y(^1 )(x)+a 0 y(x)=f(x)

Then in the following cases:

(1) f(x)anmthdegree polynomial

(a) Ifa 0   =0 – i.e. the equation contains a non-derivative term,

then we choose a particular solution of the form

yp=Lmxm+Lm− 1 xm−^1 +···+L 1 x+L 0

(b) If a 0 =0 so that there is no non-derivative term in the

differential equation and if the lowest order of derivative is

r, then choose

yp=Lmxm+r+Lm− 1 xm+r−^1 +···+L 1 xr+^1 +L 0 xr

(2) f(x)of the form sinωxor cosωx

(a) If sinωxor cosωxdo not occur in the complementary function,

then take a particular solution of the form

yp=Lcosωx+Msinωx

(b) If the complementary function contains terms of the form

xrsinωx or xrcosωx for r= 0 , 1 ,...,m,takeyp to be

the form

Lxm+^1 cosωx+Mxm+^1 sinωx

(3) f(x)an exponential function,f(x)=ekx

(a) Ifekxis not contained in the complementary function, take

yp(x)to be of the form

yp=Lekx