Calculus- Early Transcendentals, 2021a

400 Differential Equations
v ′ = 1
t 2
v = − 1 t .
Then substituting we get
u ′ e t = −v ′ te t = − 1
t 2tet
u ′ = − 1 t
u = −lnt.
The solution is Ae t + Bte t − e t lnt − e t .
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Exercises for 10.7
Find the general solution to the differential equation using variation of parameters.
Exercise 10.7.1 y ′′ + y = tanx
Exercise 10.7.2 y ′′ + y = e 2t
Exercise 10.7.3 y ′′ + 4y = secx
Exercise 10.7.4 y ′′ + 4y = tanx
Exercise 10.7.5 y ′′ + y ′ − 6y = t 2 e 2t
Exercise 10.7.6 y ′′ − 2y ′ + 2y = e t tan(t)
Exercise 10.7.7 y ′′ −2y ′ +2y = sin(t)cos(t) (This is rather messy when done by variation of parameters;
compare to undetermined coefficients.)