3x^{''}+48x=180e^{-4t}cos(4t)

Lösung

3 x^{^{''}} + 48 x = 180 e^{- 4 t} cos (4 t)

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

Schritte zur Lösung

3 x^{''} (t) + 48 x = 180 e^{- 4 t} cos (4 t)

Solve linear ODE:

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

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

y^{^{''}} + 20 y^{^{'}} + 100 y = 240 cos (10 x)

y^{^{''}} + 3 y^{^{'}} - 4 y = cos (2 x), y (0) = 1

y^{^{'''}} + 6 \cdot y^{^{''}} = x^{2}

l in e a r y^{^{''}} - 14 y^{^{'}} + 49 y = x^{2}

y^{^{''}} + 16 y^{^{'}} = x + e^{2 x}