y^{''}+6y^'+8y=t\H(t-3)

Lösung

y^{^{''}} + 6 y^{^{'}} + 8 y = t H (t - 3)

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + e^{- 2 t} \cdot \int \frac{e ^{2 t} t H ( t - 3 )}{2} d t + e^{- 4 t} \cdot \int - \frac{e ^{4 t} t H ( t - 3 )}{2} d t

Schritte zur Lösung

y^{''} (t) + 6 y^{'} (t) + 8 y = t H (t - 3)

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + e^{- 2 t} \cdot \int \frac{e ^{2 t} t H ( t - 3 )}{2} d t + e^{- 4 t} \cdot \int - \frac{e ^{4 t} t H ( t - 3 )}{2} d t

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + e^{- 2 t} \cdot \int \frac{e ^{2 t} t H ( t - 3 )}{2} d t + e^{- 4 t} \cdot \int - \frac{e ^{4 t} t H ( t - 3 )}{2} d t

\frac{1}{2} y^{^{''}} + 2 y = 4 tan (2 t) - \frac{1}{3} e^{2 t}

x^{^{''}} + 16 x = 102 e^{- 2 t} cos (4 t)

y^{^{''}} - y = 6 t + 1

y^{^{''}} - y = 6 t + 8

\frac{d ^{2} y}{d t ^{2}} + y = e^{- 2 t} sin (t)