x^{''}+2x^'+5x=10cos(3t)

Lösung

x^{^{''}} + 2 x^{^{'}} + 5 x = 10 cos (3 t)

x = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{15}{13} sin (3 t) - \frac{10}{13} cos (3 t)

Schritte zur Lösung

x^{''} (t) + 2 x^{'} (t) + 5 x = 10 cos (3 t)

Solve linear ODE:

x = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{15}{13} sin (3 t) - \frac{10}{13} cos (3 t)

x = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{15}{13} sin (3 t) - \frac{10}{13} cos (3 t)

y^{^{''}} - y^{^{'}} = 1, y (0) = 19, y^{^{'}} (0) = 1

so l v e f or d, d (x) = \frac{5}{4} x^{2} + 56 x^{4} + \frac{8}{x} - 9 x

y^{^{''}} + 8 y^{^{''}} - 9 y = - 10 e^{- x} - 425 cos (2 x)

y^{^{''}} + 16 y = sin (t), y (0) = 0, y^{^{'}} (0) = 0

9 y^{^{''}} - 6 y^{^{'}} + y = 50 sin (x)