y^{''}+4y^'+13y=cos(t)

Lösung

y^{^{''}} + 4 y^{^{'}} + 13 y = cos (t)

y = e^{- 2 t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + \frac{1}{40} sin (t) + \frac{3}{40} cos (t)

Schritte zur Lösung

y^{''} (t) + 4 y^{'} (t) + 13 y = cos (t)

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + \frac{1}{40} sin (t) + \frac{3}{40} cos (t)

y = e^{- 2 t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + \frac{1}{40} sin (t) + \frac{3}{40} cos (t)

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

\frac{d ^{2} y}{d x ^{2}} + 8 \frac{d y}{d x} + 16 y = e^{4 x}

y^{^{''}} + 16 y^{^{'}} + 10000 y = 400 sin (200 x)

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

y^{^{''}} - 4 y^{^{'}} + 4 y = \frac{- 2 e ^{2 t}}{t ^{2} + 1}