Lecture Notes in Differential Equations - Bruce E. Shapiro

182 LESSON 21. REDUCTION OF ORDER
Substitution back into the original ODE verifies that this works. Hence a
general solution of the equation is
∫
y = At + Bt t −2 e −t2 /2 dt (21.35)
Example 21.3. Find a second solution of
t 2 y ′′ − 4ty ′ + 6y = 0 (21.36)
assuming that t > 0, given that y 1 = t 2 is already a solution.
We look for a solution of the form y 2 = uy 1 = ut 2 . Differentiating
Substituting in (21.36),
y ′ 2 = u ′ t 2 + 2tu (21.37)
y ′′
2 = u ′′ t 2 + 4tu ′ + 2u (21.38)
0 = t 2 (u ′′ t 2 + 4tu ′ + 2u) − 4t(u ′ t 2 + 2tu) + 6ut 2 (21.39)
= t 4 u ′′ + 4t 3 u ′ + 2t 2 u − 4t 3 u ′ − 8t 2 u + 6t 2 u (21.40)
= t 4 u ′′ (21.41)
Since t > 0 we can never have t = 0; hence we can divide by t 4 to give
u ′′ = 0. This means u = t. Hence
y 2 = uy 1 = t 3 (21.42)
is a second solution.
Example 21.4. Show that
y 1 = sin t √
t
(21.43)
is a solution of Bessel’s equation of order 1/2
(
t 2 y ′′ + ty ′ + t 2 − 1 )
y = 0 (21.44)
4
and use reduction of order to find a second solution.
Differentiating using the product rule
y 1 ′ = d ( )
t −1/2 sin t
dt
(21.45)
= t −1/2 cos t − 1 2 t−3/2 sin t (21.46)
y ′′
1 = −t −1/2 sin t − t −3/2 cos t + 3 4 t−5/2 sin t (21.47)