Mathematics
Undergraduate
関数 x = 0 も x = e^(it) - (cos t + i sin t) も微分方程式
((d^2x)/(dt)^2) + x = 0 の解であることの証明と
x(0) = (dx(0))/(dt) = 0 の証明です。
写真のようなやり方は正しいでしょうか?
(3)
d£²
= - k
x=0 より 2階微分で
d²%
t=0と分かる。又、
d¾
at² = -(0) = 0 =) FX.
-
2 = ²t (cost + isint)
et - eit
=
=0より同様の
x=0
(₂)
dx
at = 0
x = 1 -(1+0) = 0
2+ = 2·1 - (0 + i) = 0
=0
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