3x^{''}+48x=60e^{-4t}cos(4t)

Lösung

3 x^{^{''}} + 48 x = 60 e^{- 4 t} cos (4 t)

x = c_{1} cos (4 t) + c_{2} sin (4 t) - \frac{1}{2} e^{- 4 t} sin (4 t) + \frac{1}{4} e^{- 4 t} cos (4 t)

Schritte zur Lösung

3 x^{''} (t) + 48 x = 60 e^{- 4 t} cos (4 t)

Solve linear ODE:

x = c_{1} cos (4 t) + c_{2} sin (4 t) - \frac{1}{2} e^{- 4 t} sin (4 t) + \frac{1}{4} e^{- 4 t} cos (4 t)

x = c_{1} cos (4 t) + c_{2} sin (4 t) - \frac{1}{2} e^{- 4 t} sin (4 t) + \frac{1}{4} e^{- 4 t} cos (4 t)

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = 2 e^{x}

- y^{^{''}} + y = e^{a \cdot x}, y (0) = 0, y (1) = 0

y^{^{'''}} - y^{^{''}} = x + e^{x}

y^{^{''}} + 4 y^{^{'}} + 32 y = 4 cos (4 t)

y^{^{''}} + 4 y = t^{2} + 6 e^{t}, y (0) = 0, y^{^{'}} (0) = 7