y^{''}-4y^'+53y=5e^{2t}cos(7t)

解

y^{^{''}} - 4 y^{^{'}} + 53 y = 5 e^{2 t} cos (7 t)

y = e^{2 t} (c_{1} cos (7 t) + c_{2} sin (7 t)) + \frac{5 e ^{2 t} t sin ( 7 t )}{14}

解答ステップ

y^{''} (t) - 4 y^{'} (t) + 53 y = 5 e^{2 t} cos (7 t)

線形 ODE を解く:

y = e^{2 t} (c_{1} cos (7 t) + c_{2} sin (7 t)) + \frac{5 e ^{2 t} t sin ( 7 t )}{14}

y = e^{2 t} (c_{1} cos (7 t) + c_{2} sin (7 t)) + \frac{5 e ^{2 t} t sin ( 7 t )}{14}