Lecture Notes in Differential Equations - Bruce E. Shapiro

251
Example 27.14. Find the general solution of
given that y = e t and y = e −t are solutions.
From Abel’s formula,
W (t) = exp
ty ′′′ − y ′′ − ty ′ + y = 0 (27.185)
{ ∫ }
− (−1/t)dt = exp{ln t} = t (27.186)
By direct calculation,
∣ e −t e t y ∣∣∣∣∣
W (t)=
−e −t e t y ′
(27.187)
∣ e −t e t y ′′
= e −t ∣ ∣∣∣ e t y ′
e t y ′′ ∣ ∣∣∣
+ e −t ∣ ∣∣∣ e t y
e t y ′′ ∣ ∣∣∣
+ e −t ∣ ∣∣∣ e t y
e t y ′ ∣ ∣∣∣
(27.188)
= e −t [(e t y ′′ − e t y ′ ) + (e t y ′′ − e t y) + (e t y ′ − e t y)] (27.189)
= y ′′ − y ′ + y ′′ − y + y ′ − y (27.190)
= 2y ′′ − 2y (27.191)
Setting the two expressions for the Wronskian equal to one another,
y ′′ − y = t 2
(27.192)
The method of undetermined coefficients tells us that
y = C 1 e −t + C 2 e t − 1 2 t (27.193)
is a solution ( the first two terms are y H and the third is y P ).