N-th Order Linear Differential Equations 205Factoring, we findp(r) = (r - 2)^3 = 0.
The roots of this auxiliary equation are ri = r2 = r3 = 2. Thus, one solution
of the DE (8) is Yi(x) = erix = e^2 x. However, Y2(x) = er^2 x = e^2 x = Yi(x)
and y3 ( x) = er^3 x = e^2 x = Yi ( x). That is, Y2 and y3 are identical to Yi.
Consequently, we h ave only one solution of the DE (8). So when some roots
of the auxiliary equation are real and equal, the technique of the previous
paragraph will not suffice.
A root s of the a uxiliary equation (6), p(r) = 0, is called a root of multi-
plicity k , if the unique factorization of the polynomial p(r) contains the factor
( r - s) exactly k times. The function y( x) = esx will be one member of the
linearly independent so lution set; however, fork 2: 2 we must find k -1 other
linearly independent solutions corresponding to the root s. To find the addi-
tional linearly independent solutions, we assume that there are k - 1 solutions
of the form Ym+i = xmesx, where 1 :::; m:::; k-l. We then verify that Ym+i (x)
satisfies the differential equation and that the set {Yi(x ), Y2(x), ... , Yk(x)} is
linearly independent on the interval (-oo, oo ). The functions Ym+i associated
with the repeated root s are linearly independent, because the functions 1,
x, x^2 ,... , xk-i are linearly independent. The functions Ym+i are also lin-
early independent of all solutions associated with other distinct roots of the
auxiliary equation. This is always the case, but we will not present a proof.
Returning to our example, we will verify that y2(x) = xe^2 x and y3(x) =x^2 e^2 x are solutions of the DE (8) and that Yi(x) = e^2 x, Y2(x) = xe^2 x, and
y 3 (x) = x^2 e^2 x are linearly indep endent on (-00,00). Calculating t h e first,
second, and third derivatives of Y2 and y3, we find
y?, = (1 + 2x)e^2 x and
y~ = (4 + 4x)e^2 x and y~ = (2 + 8x + 4x^2 )e^2 x
y~' = (12 + 8x )e^2 x and yr= (12 + 24x + 8x^2 )e^2 x.
Substitution of Y2 into (8), yields
y~' -6y~ +12y~ -8y2 = [(12+8x)-6(4+4x)+12(1+2x)-8x]e^2 x = O·e^2 x = 0.
Hence, y 2 is a solut ion of the DE (8). Substitution of y3 into (8), yields
y~' -6y~ +l2y~-8y 3 = [(12+24x+8x^2 )-6(2+8x+ 4x^2 )+12(2x+2x^2 ) -8x^2 ]e^2 x
= 0 · e^2 x = 0.
Thus, y 3 is a solution of the DE (8). To verify that Yi, Y2, and y3 are linearly
independent on ( -oo, oo), we examine their Wronskian at x = 0.
Yi(O)W(yi,y2,y3,0) = y~(O)
y~(O)Y2(0)
y?,(O)
y~(O)y3(0)
y~(O)
y~(O)1 0
2 1
4 400 = 2 i= 0.
2