Lecture Notes in Differential Equations - Bruce E. Shapiro

254 LESSON 27. HIGHER ORDER EQUATIONS
Example 27.15. Solve y ′′′ + y ′ = tan t using variation of parameters.
The characteristic equation is 0 = r 3 + r = r(r + i)(r − i); hence a fundamental
set of solutions are y 1 = 1, y 2 = cos t, and y 3 = sin t. From either
Abel’s formula or a direct calculation, W (t) = 1, since
y P = y 1
∫
t
W =
∣
W 1 (s)f(s)
W (s)a 3 (s) ds+y 2
1 cos t sin t
0 − sin t cos t
0 − cos t − sin t
∫
t
W 2 (s)f(s)
W (s)a 3 (s) ds+y 3
∣ = 1 (27.207)
∫
W 3 (s)f(s)
ds (27.208)
W (s)a 3 (s)
where a 3 (s) = 1, f(s) = tan s, and
0 cos t sin t
W 1 =
0 − sin t cos t
∣ 1 − cos t − sin t ∣ = 1 (27.209)
1 0 sin t
W 2 =
0 0 cos t
= − cos t (27.210)
∣ 0 1 − sin t ∣ 1 cos t 0
W 3 =
0 − sin t 0
= − sin t (27.211)
∣ 0 − cos t 1 ∣
Therefore
∫
y P =
t
∫
∫
tan sds − cos t cos s tan sds − sin t sin s tan sds (27.212)
t
t
Integrating the first term and substituting for the tangent in the third term,
∫
∫
sin 2 s
y P = − ln |cos t| − cos t sin sds − sin t ds (27.213)
cos s
t
The second integral can not be integrated immediately, and the final integral
can be solved by substituting sin 2 s = 1 − cos 2 s
∫
y P =− ln |cos t| + cos 2 1 − cos 2 s
t − sin t
ds (27.214)
cos s
Since the first term (the constant) is a solution of the homogeneous equation,
we can drop it from the particular solution, giving
and a general solution of
t
y P = ln |cos t| − sin t ln |sec t + tan t| (27.215)
y = ln |cos t| − sin t ln |sec t + tan t| + C 1 + C 2 cos t + C 3 sin t. (27.216)
t
t