Lecture Notes in Differential Equations - Bruce E. Shapiro

181
The method of reduction of order is more generally valid then for equations
with constant coefficient which we already know how to solve. It is usually
more practical to repeat the derivation rather than using equation (21.14),
which is difficult to memorize.
Example 21.2. Find a second solution to
using the observation that y 1 = t is a solution.
We look for a solution of the form
Differentiating,
y ′′ + ty ′ − y = 0 (21.21)
y 2 = y 1 u = tu (21.22)
Hence from (21.21)
y 2 ′ = tu ′ + u (21.23)
y 2 ′′ = tu ′′ + 2u ′ (21.24)
tu ′′ + 2u ′ + t 2 u ′ + tu − tu = 0 (21.25)
tu ′′ + 2u ′ + t 2 u ′ = 0 (21.26)
tz ′ + (2 + t 2 )z = 0 (21.27)
where z = u ′ . Rearranging and separating variables in z,
∫ 1
z
1 dz + t2
= −2 = − 2 z dt t t − t (21.28)
∫ ∫
dz 2
dt dt = − t dt − tdt (21.29)
ln z = −2 ln t − 1 2 t2 (21.30)
z = 1 t 2 e−t2 /2
(21.31)
Therefore
or
du
dt = 1 /2
t 2 e−t2 (21.32)
∫
u(t) = t −2 e −t2 /2 dt (21.33)
Thus a second solution is
∫
y 2 = tu = t
t −2 e −t2 /2 dt (21.34)