y^{''}+2y^'+4y=2t-3cos(2t)

Lösung

y^{^{''}} + 2 y^{^{'}} + 4 y = 2 t - 3 cos (2 t)

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + \frac{t}{2} - \frac{1}{4} - \frac{3}{4} sin (2 t)

Schritte zur Lösung

y^{''} (t) + 2 y^{'} (t) + 4 y = 2 t - 3 cos (2 t)

Solve linear ODE:

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + \frac{t}{2} - \frac{1}{4} - \frac{3}{4} sin (2 t)

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + \frac{t}{2} - \frac{1}{4} - \frac{3}{4} sin (2 t)

3 y^{^{''}} - 9 y^{^{'}} = sin (2 x)

y^{^{''}} + 9 y = 3, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 0.01 \cdot (20 - y) = 0

y^{^{''}} + 4 y = 8 e^{2 t}, y (0) = 1, y^{^{'}} (0) = - 2

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} = 2 x + 5 - e^{- 2 x}