S.H.I.
Оракул
(68482)
1 месяц назад
y'' = tan x ⋅ 1 / cos²x
y(0) = 1/2
y′(0) = 0
∫y'' dx = ∫tan x / cos²x dx + C
= ∫sin x / cos³x dx + C
= -∫du / u³
= 1 / (2u²) + C
= 1 / (2 cos²x) + C
y′ = 1 / (2 cos²x) + C
∫y′ dx = ∫(1 / 2 cos²x + C) dx + D
= 1/2 tan x + Cx + D
y(0) = 1/2 ⇒ D = 1/2
y′(0) = 0 ⇒ 1/2 + C = 0 ⇒ C = -1/2
y(x) = 1/2 tan x - 1/2 x + 1/2
y(π/4) = 1 - π / 8