Lecture Notes in Differential Equations - Bruce E. Shapiro

180 LESSON 21. REDUCTION OF ORDER
which is separable in z.
1 dz
z dt = −2y′ 1 + py 1
= −2 y′ 1
− p (21.9)
y 1 y 1
∫ y
′
∫
ln z = −2 1
− pdt (21.10)
y 1
∫
= −2 ln y 1 − pdt (21.11)
z = 1 ( ∫ )
y1 2(t)exp − p(t)dt
(21.12)
Since z = g ′ (t),
g(t) =
∫ ( ( ∫
1
y1 2(t)exp −
))
p(t)dt dt (21.13)
and since y 2 = gy 1 , the method of reduction of order gives
∫ ( ( ∫
1
y 2 (t) = y 1 (t)
y1 2(t)exp −
Example 21.1. Use (21.14) to find a second solution to
given that one solution is y 1 = e 2t .
))
p(t)dt dt (21.14)
y ′′ − 4y ′ + 4y = 0 (21.15)
Of course we already know that since the characteristic equation is (r−2) 2 =
0, the root r = 2 is repeated and hence a second solution is y 2 = te 2t . We
will now derive this solution with reduction of order.
From equation (21.14), using p(t) = −4,
∫ ( ( ∫ ))
1
y 2 (t) = y 1 (t)
y1 2(t)exp − p(t)dt dt (21.16)
∫ (∫ ))
= e
(e 2t −4t exp 4dt dt (21.17)
= e 2t ∫ (e −4t e 4t) dt (21.18)
= e 2t ∫
dt (21.19)
= te 2t (21.20)