x^{''}+x^'+12x=40cos(3t)

Lösung

x^{^{''}} + x^{^{'}} + 12 x = 40 cos (3 t)

x = e^{- \frac{t}{2}} (c_{1} cos (\frac{47 t}{2}) + c_{2} sin (\frac{47 t}{2})) + \frac{20}{3} sin (3 t) + \frac{20}{3} cos (3 t)

Schritte zur Lösung

x^{''} (t) + x^{'} (t) + 12 x = 40 cos (3 t)

Solve linear ODE:

x = e^{- \frac{t}{2}} (c_{1} cos (\frac{47 t}{2}) + c_{2} sin (\frac{47 t}{2})) + \frac{20}{3} sin (3 t) + \frac{20}{3} cos (3 t)

x = e^{- \frac{t}{2}} (c_{1} cos (\frac{47 t}{2}) + c_{2} sin (\frac{47 t}{2})) + \frac{20}{3} sin (3 t) + \frac{20}{3} cos (3 t)

y^{^{''}} - 3 y^{^{'}} = e^{3 x}

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 e^{t}

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = 9 e^{- x}

y^{^{''}} + 4 y = - 20 sin (2 t)

y^{^{''}} - 6 y^{^{'}} + 9 y = 2 e^{x}