x^{''}+2.5x^'+50x=1.5cos(t)

Lösung

x^{^{''}} + 2.5 x^{^{'}} + 50 x = 1.5 cos (t)

x = e^{- \frac{5 t}{4}} (c_{1} cos (\frac{5 31 t}{4}) + c_{2} sin (\frac{5 31 t}{4})) + 0.00155 \dots sin (t) + \frac{294}{9629} cos (t)

Schritte zur Lösung

x^{''} (t) + 2.5 x^{'} (t) + 50 x = 1.5 cos (t)

Solve linear ODE:

x = e^{- \frac{5 t}{4}} (c_{1} cos (\frac{5 31 t}{4}) + c_{2} sin (\frac{5 31 t}{4})) + 0.00155 \dots sin (t) + \frac{294}{9629} cos (t)

x = e^{- \frac{5 t}{4}} (c_{1} cos (\frac{5 31 t}{4}) + c_{2} sin (\frac{5 31 t}{4})) + 0.00155 \dots sin (t) + \frac{294}{9629} cos (t)

y^{^{''}} - 4 y = 1 - cos (t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + y = b x

y^{^{''}} - 8 y^{^{'}} + 15 y = 9 x e^{2 x}

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

2 y^{^{''}} - 7 y^{^{'}} + 3 y = 5 e^{2 x}