Lecture Notes in Differential Equations - Bruce E. Shapiro

209
Example 24.3. Solve the initial value problem
⎫
t 2 y ′′ − 2y = 2t⎪⎬
y(1) = 0
⎪⎭
y ′ (1) = 1
(24.41)
given the observation that y = t 2 is a homogeneous solution.
First, we find a second homogeneous solution using reduction of order. By
Abel’s formula, since there is no y ′ term, the Wronskian is a constant:
W (t) = e − ∫ 0dt = C (24.42)
A direct calculation of the Wronskian gives
W (t) =
∣ t2 y 2
2t y 2
′ ∣ = t2 y 2 ′ − 2ty 2 (24.43)
Setting the two expressions for the Wronskian equal to one another gives
t 2 y ′ − 2ty = C (24.44)
where we have omitted the subscript. This is a first order linear equation
in y;putting it into standard form,
y ′ − 2 t y = C t 2 (24.45)
An integrating factor is µ = e ∫ (−2/t)dt = e −2 ln t = t −2 , hence
Integrating both sides of the equation over t,
Multiplying through by t 2 ,
d y
dt t 2 = Ct−4 (24.46)
y
t 2 = −C 3 t−3 + C 1 (24.47)
y = C 1 t 2 + C 2
t
(24.48)
where C 2 = C/ − 3. Since y 1 = t 2 , we conclude that y 2 = 1/t. This gives
us the entire homogeneous solution.