N-th Order Linear Differential Equations 209
Furthermore, if the complex number a + i{3 where {3 ~ 0 and its complex
conjugate a -i{3 are both roots of multiplicity k > 0 of the auxiliary equation
associated with a homogeneous linear differential equation, then 2k linearly
independent, real solutions corresponding to these two roots are
Y1 = e°'x cos {3x,
Y3 = xe°'x cos {3x,
Y2 = e°'x sin {3x,
Y4 = xe°'x sin {3x,
and the general solution of the differential equation will include the linear
combination C1Y 1 + C2Y2 + · · · + c2k-1Y2k-1 + C2kY2k· This result is important
since the conjugate root theorem states that if the coefficients of an n-th
degree polynomial are all real, then complex roots occur in conjugate pairs.
So if the coefficients of an n-th order linear differential equation are all real
constants, then complex roots of the associated a uxilia ry equation will always
occur in conjugate p airs, all n linearly independent solutions may be written as
real solutions, and the general solution can be written as a linear combination
of real solut ions. It should be noted that if a ny of the constant coefficients
an, an-1, ... , ai, ao of the differential equation (hence, auxiliary equation)
is complex, then it is no longer true that the complex conjugate of a complex
root will also be a root. In this case, the corresponding general solution will
contain some complex components.
EXAMPLE 3 Real General Solution of a Homogeneous Linear
Differential Equation
Find the real general solut ion of the fifth order linear homogeneous differ-
ential equation
y(^5 ) - lly(^4 ) + 50y(^3 ) - 94y(^2 ) + 13y(l) + 16 9y = 0.
SOLUTION
Using the computer program POLYRTS, we find the roots of the associated
a uxili ary equation
r^5 - llr^4 + 50r^3 - 94r^2 + 13r^1 + 169 = 0
are - 1, 3 + 2i, 3 + 2i, 3 - 2i , and 3 - 2i. Corresponding to the real root -1,
is the real solution
Y1=e-x.
Corresponding to the double complex root 3 + 2i and the double complex
conjugate root 3 - 2i, a re the four linearly independent real solutions
y 2 = e^3 x cos 2x, y3 = e^3 x sin 2x, y4 = xe^3 x cos 2x, and y5 = xe^3 x sin 2x.