Understanding Engineering Mathematics

For the PI you might be tempted to tryyp=Lcos 3x, but in fact we need both cos 3x
and sin 3xterms because of the first order derivative on the left-hand side. We therefore try

yp=Lcos 3x+Msin 3x

yp′=− 3 Lsin 3x+ 3 Mcos 3x

yp′′=− 9 Lcos 3x− 9 Msin 3x

substituting in the DE gives, on collecting cos and sin terms:

(− 7 L+ 6 M)cos 3x+(− 6 L− 7 M)sin 3x≡2cos3x

So
− 6 L− 7 M= 0
− 7 L+ 6 M= 2

whence

L=−

14

85

,M=

12

85

and the PI is

yp=−

14

85

cos 3x+

12

85

sin 3x

and the GS is

y=e−x(Acosx+Bsinx)−

14

85

cos 3x+

12

85

sin 3x

(again, check back in the equation).

Problem 15.20

Find the general solution to the equation

y′′Y 4 y=e−xcosx

The CF is
yc=Acos 2x+Bsin 2x

For the PI choose

yp=Le−xcosx+Me−xsinx

Then, as an exercise in differentiation by the product rule (234
➤
), you can check that

yp′′=e−x( 2 Lsinx− 2 Mcosx)

Substituting in the equation gives, after simplification

e−x[( 4 L− 2 M)cosx+( 2 L+ 4 M)sinx]≡e−xcosx