y^'+(2y)/(t-2)=(sin(t))/(t-2)

解

y^{^{'}} + \frac{2 y}{t - 2} = \frac{sin ( t )}{t - 2}

y = - \frac{cos ( t )}{t - 2} + \frac{sin ( t )}{t ^{2} - 4 t + 4} + \frac{c _{1}}{t ^{2} - 4 t + 4}

解答ステップ

y^{'} (t) + \frac{2 y}{t - 2} = \frac{sin ( t )}{t - 2}

線形 ODE を解く:

y = - \frac{cos ( t )}{t - 2} + \frac{sin ( t )}{t ^{2} - 4 t + 4} + \frac{c _{1}}{t ^{2} - 4 t + 4}

y = - \frac{cos ( t )}{t - 2} + \frac{sin ( t )}{t ^{2} - 4 t + 4} + \frac{c _{1}}{t ^{2} - 4 t + 4}