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

Lösung

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

x = e^{- \frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2})) + \frac{1}{4} sin (2 t) - \frac{1}{4} cos (2 t)

Schritte zur Lösung

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

Solve linear ODE:

x = e^{- \frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2})) + \frac{1}{4} sin (2 t) - \frac{1}{4} cos (2 t)

x = e^{- \frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2})) + \frac{1}{4} sin (2 t) - \frac{1}{4} cos (2 t)

\frac{d ^{2} y}{d x ^{2}} - 16 y = 9.6 e^{4 x} + 30 e^{x}

y^{^{'''}} = 3 y^{^{''}} + 9 (e^{2 x}) + 5 (cos (x))

y^{^{''}} + 7 y^{^{'}} + 10 y = 7 u

y^{^{''}} + 4 y = cos (x), y (0) = 0, y^{^{'}} (0) = 0

2 y^{^{''}} + y^{^{'}} = 3 x^{2} + 10 x