Recall the method for solving an nth-order homogenous differential equation by finding roots of an auxiliary equation.
any(n) +
an − 1y(n − 1) +
+
a1y′ +
a0y = 0
Rewrite this as a polynomial in
m, replacing the
jth derivative with
mj.
anmn +
an − 1mn − 1 +
+
a1m +
a0 = 0
Then for every root
mi
of multiplicity
k of this polynomial, the general solution for the equation must contain the following linear combination.
c1em1x +
c2xem1x +
+
ck xk − 1em1x
We are given the following third-order homogeneous differential equation.
Therefore, the auxiliary equation is a third-degree polynomial, which can be factored as follows.
m3 + 12m2 + 36m | = | 0 |
m(m2 + 12m + 36) | = | 0 |
m(m + 6)m + 6
6
| = | 0 |
Solving for
m, the roots of the auxiliary equation are
m1 =
0
0
with multiplicity 1, and
m2 =
-6
-6
with multiplicity 2.