3x^{''}+48x=153e^{-2t}cos(4t)

Lösung

3 x^{^{''}} + 48 x = 153 e^{- 2 t} cos (4 t)

x = c_{1} cos (4 t) + c_{2} sin (4 t) - 3 e^{- 2 t} sin (4 t) + \frac{3}{4} e^{- 2 t} cos (4 t)

Schritte zur Lösung

3 x^{''} (t) + 48 x = 153 e^{- 2 t} cos (4 t)

Solve linear ODE:

x = c_{1} cos (4 t) + c_{2} sin (4 t) - 3 e^{- 2 t} sin (4 t) + \frac{3}{4} e^{- 2 t} cos (4 t)

x = c_{1} cos (4 t) + c_{2} sin (4 t) - 3 e^{- 2 t} sin (4 t) + \frac{3}{4} e^{- 2 t} cos (4 t)

y^{^{''}} - 9 y = 4 e^{3 t}, y (0) = 0, y^{^{'}} (0) = - 1

y^{^{''}} - 3 y^{^{'}} + 2 y = 2 cos^{2} (4 x)

y^{^{'''}} + 2 y^{^{''}} + y^{^{'}} = 2 - 24 e^{x} + 40 e^{5 x}

y^{^{''}} + 2 y = cos (2) t + sin (2) t

y^{^{''}} + 25 y = 7 sec (5 x)