y^{''}+40y^'+625y=100cos(10t)

Lösung

y^{^{''}} + 40 y^{^{'}} + 625 y = 100 cos (10 t)

y = e^{- 20 t} (c_{1} cos (15 t) + c_{2} sin (15 t)) + \frac{64}{697} sin (10 t) + \frac{84}{697} cos (10 t)

Schritte zur Lösung

y^{''} (t) + 40 y^{'} (t) + 625 y = 100 cos (10 t)

Solve linear ODE:

y = e^{- 20 t} (c_{1} cos (15 t) + c_{2} sin (15 t)) + \frac{64}{697} sin (10 t) + \frac{84}{697} cos (10 t)

y = e^{- 20 t} (c_{1} cos (15 t) + c_{2} sin (15 t)) + \frac{64}{697} sin (10 t) + \frac{84}{697} cos (10 t)

2 y^{^{''}} - y^{^{'}} - 2 y = 2 e^{- t}

y^{^{''}} + 4 y = 20 e^{t}, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - y = 4 x + 3

y^{^{''}} - 15 y^{^{'}} + 54 y = 39 + 54 x

x^{^{''}} + 3 x^{^{'}} + 2 x = 1, x (0) = 0, x^{^{'}} (0) = 0