(dx)/(dt)=sin(2pi(t+3)),x(3)=1

解

\frac{d x}{d t} = sin (2 π (t + 3)), x (3) = 1

x = - \frac{1}{2 π} cos (2 π (t + 3)) + 1 + \frac{1}{2 π}

解答ステップ

\frac{d x}{d t} = sin (2 π (t + 3))

代用

\frac{d x}{d t}

と

x^{'} (t)

x^{^{'}} (t) = sin (2 π (t + 3))

f^{^{'}} (x) = g (x)

であれば

f (x) = \int g (x) d x

x = \int sin (2 π (t + 3)) d t

\int sin (2 π (t + 3)) d t = - \frac{1}{2 π} cos (2 π (t + 3)) + c_{1}

x = - \frac{1}{2 π} cos (2 π (t + 3)) + c_{1}

初期条件を適用する:

x = - \frac{1}{2 π} cos (2 π (t + 3)) + 1 + \frac{1}{2 π}

x = - \frac{1}{2 π} cos (2 π (t + 3)) + 1 + \frac{1}{2 π}