Lecture Notes in Differential Equations - Bruce E. Shapiro

185
Example 21.5. Find a second solution to the differential equation
given the observation that y 1 (t) = 1 is a solution.
Since p(t) = 10/t, Abel’s formula gives
ty ′′ + 10y ′ = 0 (21.63)
W (t) = e − ∫ (10/t)dx = t −10 (21.64)
By direct calculation,
W (t) =
∣ y ∣ 1 y 2
∣∣∣ y 1 ′ y 2
′ ∣ = 1 y 2
0 y 2
′ ∣ = y′ 2 (21.65)
Equating the two expressions for W,
y ′ 2 = t −10 (21.66)
Therefore, y 2 = −(1/9)t −9 ; the general solution to the homogeneous equation
is
y = C 1 y 1 + C 2 y 2 = C 1 + C 2 t −9 . (21.67)
Example 21.6. Find a fundamental set of solutions to
given the observation that y 1 = t is one solution.
t 2 y ′′ + 5ty ′ − 5y = 0 (21.68)
Calculating the Wronskian directly,
W (t) =
∣ t y 2
1 y 2
′ ∣ = ty′ 2 − y 2 (21.69)
From Abel’s Formula , since p(t) = a 1 (t)/a 2 (t) = 5/t,
W (t) = e − ∫ (5/t)dt = t −5 (21.70)
Equating the two expressions and putting the result in standard form,
An integrating factor is
Therefore
y ′ 2 − (1/t)y 2 = t −6 (21.71)
µ(t) = e ∫ (−1/t)dx = 1/t (21.72)
∫
[ ] t
y 2 ′ = t (1/t)t −6 −6
dt = t = 1 6 5 t−5 (21.73)
The fundamental set of solutions is therefore {t, t −5 }.