36600 MA LECTURE NOTES: WEDNESDAY, FEBRUARY 25
3
Repeated Roots
Review. We recap what we know so far. Consider the constant coecient dierential equation
a y + b y + c y = 0
We know that we can nd solutions by considering the characteristic equation
a r2 + b r + c = 0.
Say that roots are r1 and r2 . Using the Quadratic Formula, we can explicitly write
b
b2 4 a c
b
b2 4 a c
r1 =
+
and
r2 =
.
2a
2a
2a
2a
If the discriminant b2 4 a c > 0, we have seen that a fundamental set of solutions is given by
y1 (t) = er1 t
and
y2 (t) = er2 t .
If the discriminant b2 4 a c < 0, w have seen that a fundamental set of solutions is given by
y1 (t) = et cos t
and
y2 (t) = et sin t.
What if the discriminant b2 4 a c = 0 i.e., we have repeated roots? How do we nd all solutions?
We consider an example for motivation. Consider the dierential equation
y = 0.
Then the characteristic equation r2 = 0 has the roots r1 = r2 = 0. We know that one solution if y1 (t) =
er1 t = 1. We try a dierent method in order to nd all solutions to the equation. We see
y that
d = y = 0
=
y = c1
dt
for some constant c1 . This is a rst order linear dierential equation, which we can solve:
y (t) = c1 t + c2
for constants c1 and c2 . This is the general solution for the dierential equation.
dAlemberts Method. We now return to the general situation. Say that we wish to nd the general
solution to the dierential equation
a y + b y + c y = 0
where b2 4 a c = 0. First note that the roots of the characteristic polynomial a r2 + b r + c = 0 are
b
b2 4 a c
b
r1 = r 2 =
= .
2a
2a
2a
In particular, one solution to the dierential equation is
y1 (t) = er1 t = e(b/2a)t .
Say that y = y (t) is the general solution for the dierential equation. We assume that this solution can be
written in the form
y (t) = v (t) y1 (t)
for some function v = v (t) to be found. We have the derivatives
y = v er t
This gives the expression
y = (v + r v ) ert
y = v + 2 r v + r2 v ert
a y + b y + c y = a v + 2 r v + r2 v ert + b (v + r v ) ert + c v ert
= a v ert + (2 a r + b) v ert + a r2 + b r + c v ert
= a ert v .