y^{''}+9y=e^{2t}\H(t-1)

解

y^{^{''}} + 9 y = e^{2 t} H (t - 1)

y = c_{1} cos (3 t) + c_{2} sin (3 t) + cos (3 t) \cdot \int - \frac{sin ( 3 t ) e ^{2 t} H ( t - 1 )}{3} d t + sin (3 t) \cdot \int \frac{cos ( 3 t ) e ^{2 t} H ( t - 1 )}{3} d t

解答ステップ

y^{''} (t) + 9 y = e^{2 t} H (t - 1)

線形 ODE を解く:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + cos (3 t) \cdot \int - \frac{sin ( 3 t ) e ^{2 t} H ( t - 1 )}{3} d t + sin (3 t) \cdot \int \frac{cos ( 3 t ) e ^{2 t} H ( t - 1 )}{3} d t

y = c_{1} cos (3 t) + c_{2} sin (3 t) + cos (3 t) \cdot \int - \frac{sin ( 3 t ) e ^{2 t} H ( t - 1 )}{3} d t + sin (3 t) \cdot \int \frac{cos ( 3 t ) e ^{2 t} H ( t - 1 )}{3} d t