y^{''}-4y^'+13y=e^{2t}*sin(3t),y(0)=4

解

y^{^{''}} - 4 y^{^{'}} + 13 y = e^{2 t} \cdot sin (3 t), y (0) = 4

y = e^{2 t} (4 cos (3 t) + c_{2} sin (3 t)) - \frac{e ^{2 t} t cos ( 3 t )}{6}

解答ステップ

y^{''} (t) - 4 y^{'} (t) + 13 y = e^{2 t} \cdot sin (3 t)

線形 ODE を解く:

y = e^{2 t} (4 cos (3 t) + c_{2} sin (3 t)) - \frac{e ^{2 t} t cos ( 3 t )}{6}

y = e^{2 t} (4 cos (3 t) + c_{2} sin (3 t)) - \frac{e ^{2 t} t cos ( 3 t )}{6}