y^{''}+8y^'+12=1-\H(t-3)

解

y^{^{''}} + 8 y^{^{'}} + 12 = 1 - H (t - 3)

y = c_{1} + c_{2} e^{- 8 t} + \int \frac{- H ( t - 3 ) - 11}{8} d t + e^{- 8 t} \cdot \int - \frac{- H ( t - 3 ) - 11}{8 e ^{- 8 t}} d t

解答ステップ

y^{''} (t) + 8 y^{'} (t) + 12 = 1 - H (t - 3)

線形 ODE を解く:

y = c_{1} + c_{2} e^{- 8 t} + \int \frac{- H ( t - 3 ) - 11}{8} d t + e^{- 8 t} \cdot \int - \frac{- H ( t - 3 ) - 11}{8 e ^{- 8 t}} d t

y = c_{1} + c_{2} e^{- 8 t} + \int \frac{- H ( t - 3 ) - 11}{8} d t + e^{- 8 t} \cdot \int - \frac{- H ( t - 3 ) - 11}{8 e ^{- 8 t}} d t