x^{''}+2x^'+7x=6cos(2t)

Lösung

x^{^{''}} + 2 x^{^{'}} + 7 x = 6 cos (2 t)

x = e^{- t} (c_{1} cos (6 t) + c_{2} sin (6 t)) + \frac{24}{25} sin (2 t) + \frac{18}{25} cos (2 t)

Schritte zur Lösung

x^{''} (t) + 2 x^{'} (t) + 7 x = 6 cos (2 t)

Solve linear ODE:

x = e^{- t} (c_{1} cos (6 t) + c_{2} sin (6 t)) + \frac{24}{25} sin (2 t) + \frac{18}{25} cos (2 t)

x = e^{- t} (c_{1} cos (6 t) + c_{2} sin (6 t)) + \frac{24}{25} sin (2 t) + \frac{18}{25} cos (2 t)

y^{^{''}} + 4 y^{^{'}} + 4 y = (x + 1) e^{2 x}

y^{^{'''}} - y^{^{''}} = 4 x + 2 x e^{- 2 x}

y^{^{''}} - 8 y^{^{'}} + 16 y = 4 e^{4 x} ln (x)

y^{^{''}} - 3 y^{^{'}} + 2 y = \frac{1}{t}, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} - 2 y^{^{'}} + y = x^{3} + 9 x