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Through this substitution, the above ODE reduces to an algebraic equation,
and the solution of this algebraic equation then reduces to finding the roots
of the polynomial:

as + a s n−1 +…+ a s a+ = 0 (6.47)
n
n n−1 1 0
We learned in Chapter 5 the MATLAB command for finding these roots,
when needed. Now, using the superposition principle, and assuming all the
roots are distinct, the general solution of the homogeneous differential equa-
tion is given by:

y = c exp( s t +) c exp( s t +…+) c exp( s t) (6.48)
homog. 1 1 2 2 n n
where s , s , …, s are the above roots and c , c , …, c are constant determined
2
1
2
n
n
1
from the initial conditions of the solution and all its derivatives to order n – 1.
NOTE In the case that two or more of the roots are equal, it is easy to verify
that the solution of the homogeneous ODE includes, instead of a constant
multiplied by the exponential term corresponding to that root, a polynomial
multiplying the exponential function. The degree of this polynomial is (m – 1)
if m is the degeneracy of the root in question.
Example 6.4
Find the transient solutions to the second-order differential equation.

2
dy dy
a 2 + b + cy = 0 (6.49)
dt dt
Solution: The characteristic polynomial associated with this ODE is the sec-
ond-degree equation given by:

as + bs c+ = 0 (6.50)
2

−± b − 4 ac
2
b
The roots of this equation are s =
±
a 2
The nature of the solutions is very dependent on the sign of the descriminant
(b – 4ac):
2
2
• If b – 4ac > 0, the two roots are distinct and real. Call these roots
α and α ; the solution is then:
2
1
y = c exp(α t) + c exp(α t) (6.51)
homog. 1 1 2 2
© 2001 by CRC Press LLC