2. HOMOGENEOUS LINEAR EQUATIONS: THE GENERAL SOLUTION
83
b
Note, if the discriminant = b2 4ac = 0, then r1 = r2 =
, giving the
2a
characteristic equation a double or repeated root of multiplicity 2.
We have:
Lemma. ert is a solution of (⇤) () r is a root of ar2 + br + c = 0.
Note. The same is true for Cert.
But are there still other solutions?
Note that
r1 + r2 =
Then
y 00
(y 0
Now let x = y 0
b
c
and r1r2 = .
a
a
ay 00 + by 0 + cy = 0 ()
b
c
y 00 + y 0 + y = 0 ()
a
a
(r1 + r2)y 0 + r1r2y = 0 ()
r2xy)0
r1(y 0
r2y) = 0.
r2y. Then
x0
r1x = 0,
which has the general solution x = k1er1t for k1 2 R. Substituting this solution
into x = y 0 r2y,
y0
r2e r2ty = k1er1te r2t ()
d h r2t i
(#)
e y = k1e(r1 r2)t.
dt
Suppose r1 6= r2. Then, integrating,
Z
k1
e r2ty = k1e(r1 r2)t dt + k2 =
e(r1 r2)t + k2 ()
r1 r2
e
r2 t 0
r2y = k1er1t ()
y