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

Lösung

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

Schritte zur Lösung

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

Solve linear ODE:

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

y^{^{''}} - 2 y^{^{'}} + y = t \cdot e^{t} + 4, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} - y^{^{'}} - 6 y = 33 sin (3 x)

y^{^{'''}} + y^{^{''}} - 2 y = e^{3 x} (9 + 67 x + 17 x^{2})

y^{^{''}} - 8 y^{^{'}} + 80 y = - 2 sin (3 t)

0.25 y^{^{''}} + 0.25 y^{^{'}} + 16 y = 3 cos (2 t)