Understanding Engineering Mathematics

or with C=A+BandD=A−jB

y=eαx(Ccosβx+Dsinβx)

Note: Although we usually prefer such a real form of the solution, particularly in actual
physical applications, there are times when the complex exponential form is in fact
most convenient – for example when we are using phasors in alternating current elec-
trical work.
We now work through a number of problems to illustrate the above results. As always,
you will benefit greatly by checking the solutions obtained by substituting back into
the DE.

Problem 15.14

Find values oflfor whichy=elxsatisfies the DE

y′′Y 3 y′Y 2 y= 0

and hence determine its general solution. Find the particular solution satis-

fying the initial conditionsy. 0 /=0,y′. 0 /=1.

We have

y′=λeλx,y′′=λ^2 eλx

Substituting into the equation gives

(λ^2 + 3 λ+ 2 )eλx= 0

or the auxiliary equation:

λ^2 + 3 λ+ 2 = 0

(λ+ 1 )(λ+ 2 )= 0

So
λ=− 1 ,− 2

We therefore have two solutions to the DE

e−x,e−^2 x

The general solution is then

y=Ae−x+Be−^2 x

containing two arbitrary constantsA,B, as we might expect for a second order DE. To find

them we apply the given initial conditions (remember thate^0 =1and

d(eax)

dx

=aeax):

y( 0 )=A+B= 0

y′( 0 )=−A− 2 B= 1