Page 72 - Advanced engineering mathematics

52 CHAPTER 2 Linear Second-Order Equations

and
a 2
−ax/2
−ax/2
y = u e − au e + ue −ax/2 .

4
Substitute these into equation (2.3) to get
a 2
−ax/2
−ax/2
u e − au e + ue −ax/2
4
a
−ax/2
+ au e − a ue −ax/2 + bue −ax/2
2
2
a

= e −ax/2 u − b − = 0.
4
2
Since b − a /4 = 0 in this case and e −ax/2 never vanishes, this equation reduces to

u = 0.
This has solutions u(x) = cx + d with c and d as arbitrary constants. Therefore, any function
y =(cx +d)e −ax/2 is also a solution of equation (2.3) in this case. Since we need only one solution
that is linearly independent from e −ax/2 , choose c =1 and d =0 to get the second solution xe −ax/2 .
The general solution in this repeated roots case is
y = c 1 e −ax/2 + c 2 xe −ax/2 .
This is often written as y = e −ax/2 (c 1 + c 2 x).
It is not necessary to repeat this derivation every time we encounter the repeated root case.
Simply write one solution e −ax/2 , and a second, linearly independent solution is xe −ax/2 .

EXAMPLE 2.7
We will solve y + 8y + 16y = 0. The characteristic equation is

2
λ + 8λ + 16 = 0
with repeated root λ =−4. The general solution is
y = c 1 e −4x + c 2 xe −4x .

Case 3: Complex Roots
2
The characteristic equation has complex roots when a −4b<0. Because the characteristic equa-
tion has real coefficients, the roots appear as complex conjugates α + iβ and α − iβ in which α
can be zero but β is nonzero. Now the general solution is
y = c 1 e (α+iβ)x + c 2 e (α−iβ)x

or
y = e αx
c 1 e iβx + c 2 e −iβx . (2.5)
This is correct, but it is sometimes convenient to have a solution that does not involve complex
numbers. We can find such a solution using an observation made by the eighteenth century Swiss
mathematician Leonhard Euler, who showed that, for any real number β,
e iβx = cos(βx) + i sin(βx).

Problem 24 suggests a derivation of Euler’s formula. By replacing x with −x,wealsohave
e −iβx = cos(βx) − i sin(βx).

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October 14, 2010 14:12 THM/NEIL Page-52 27410_02_ch02_p43-76