Understanding Engineering Mathematics

Or, cancellinge−xand equating coefficients of cosxand sinx

4 L− 2 M= 1

2 L+ 4 M= 0

These give

L=

1

5

,M=−

1

10

So finally, the GS isy=Acos 2x+Bsin 2x+

1

5

e−xcosx−

1

10

e−xsinx

Two points to note:

• The above method of undetermined coefficients is not the only
  method – there are others that are more powerful in some cases – D-
  operators, complex variable methods, Laplace transform, etc. But it is
  systematic and routine.


• It doesn’t always work in such a straightforward way.

To see how the method can break down, and what to do about it, try the following
problem.

Problem 15.21

Investigate the particular integral for the following equation

y′′−y′− 2 y=e−x

Tryingy=Le−xfor a PI, as we might be tempted, gives

‘0=e−x’

and clearly does not work. In fact in this case the reason a trial function similar to the right-
hand side won’t work is that it is already a solution of the corresponding homogeneous
equation – that is,it is part of the complementary function. This is when the method
breaks down – when the right-hand side is part of the CF. And that’s why we always
evaluate the CF first, to check whether it does contain the right-hand side. In this case the
CF is the GS of

y′′−y′− 2 y= 0

which you can check isy=Ae−x+Be^2 x. The right-hand side is indeed included in
the CF.
So, what do we do in such cases? As usual in mathematics we do as little as
possible – that is we look for alternatives that are as close to what we have as possible.
Remember that in the case of equal roots for the auxiliary equation, when looking at the
CF, we tried – successfully – replacingeαxbyxeαx( 466

➤
)? Perhaps a similar strategy
worksthistime?

Problem 15.22

Tr yyp=Lxe−xin the differential equation of Problem 15.21, whereLis

to be determined.