Lecture Notes in Differential Equations - Bruce E. Shapiro

184 LESSON 21. REDUCTION OF ORDER
All of the terms involving u cancel out and we are left with
0 = u ′′ t 3/2 sin t + 2u ′ t 3/2 cos t (21.55)
Letting z = u ′ and dividing through by t 3/2 sin t,
Since z = u ′ this means
Since y 2 (t) = uy 1 (t),
Thus the general solution is
0 = z ′ + 2z cot t (21.56)
∫ ∫ dz
z = −2 cot tdt (21.57)
ln z = −2 ln sin t (21.58)
du
dt = 1
sin 2 t
y 2 =
y = Ay 1 + By 2 =
z = sin −2 t (21.59)
cot t sin t
√
t
=⇒ u = cot t (21.60)
= cos t √
t
(21.61)
A sin t + B cos t
√
t
(21.62)
Method for Reduction of Order. Given a first solution u(t) to
y ′′ + p(t)y ′ + q(t)y = 0, we can find a second linearly independent solution
v(t) as follows:
1. Calculate the Wronskian using Abel’s formula,
W (t) = e − ∫ p(t)dt
2. Calculate the Wronskian directly a second time as
W (t) = u ′ (t)v(t) − v ′ (t)u(t)
3. Set the two expressions equal; the result is a first order differential
equation for the second solution v(t).
4. Solve the differential equation for v(t).
5. Then general solution is then y = Au(t) + Bv(t) for some constants
A and B.