Second Order Linear Differential Equations

Case II: The characteristic equation has only one real root, r = α. 2 Then
y1(x) =e αx
and y2(x) =xe αx
are linearly independent solutions of equation (1) and
is the general solution.
y = C1 e αx + C2 xe αx
Proof: We know that y1(x) =e αx is one solution of the differential equation;
we need to find another solution which is independent of y1. Since the
characteristic equation has only one real root, α, the equation must be
r 2 + ar + b =(r − α) 2 = r 2 − 2αr + α 2 =0
and the differential equation (1) must have the form
y ′′ − 2αy ′ + α 2 y =0. (∗)
Now, z = Ce αx , C any constant, is also a solution of (∗), but z is not
independent of y1 since it is simply a multiple of y1. We replace C by a
function u which is to be determined (if possible) so that y = ue αx is a
solution of (∗). 3 Calculating the derivatives of y, we have
Substitution into (∗) gives
y = ue αx
y ′ = αue αx + u ′ e αx
y ′′ = α 2 ue αx +2αu ′ e αx + u ′′ e αx
α 2 ue αx +2αu ′ e αx + u ′′ e αx − 2α αue αx + u ′ e αx + α 2 ue αx =0.
This reduces to
u ′′ e αx = 0 which implies u ′′ = 0 since e αx =0.
Now, u ′′ = 0 is the simplest second order, linear differential equation with
constant coefficients; the general solution is u = C1 + C2x = C1 · 1+C2 · x ,
and u1(x) = 1 and u2(x) =x form a fundamental set of solutions.
Since y = ue αx , we conclude that
y1(x) =1· e αx = e αx
and y2(x) =xe αx
2
In this case, α is said to be a double root of the characteristic equation.
3
This is an application of a general method called variation of parameters. We will use the method
several times in the work that follows.
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