(D^2+4D+3)y=2xe^{3x}+3e^{3x}cos(2x)

Lösung

(D^{2} + 4 D + 3) y = 2 x e^{3 x} + 3 e^{3 x} cos (2 x)

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{e ^{3 x} x}{12} - \frac{5}{144} e^{3 x} + \frac{3}{40} e^{3 x} sin (2 x) + \frac{3}{40} e^{3 x} cos (2 x)

Schritte zur Lösung

(D^{2} + 4 D + 3) y = 2 x e^{3 x} + 3 e^{3 x} cos (2 x)

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{e ^{3 x} x}{12} - \frac{5}{144} e^{3 x} + \frac{3}{40} e^{3 x} sin (2 x) + \frac{3}{40} e^{3 x} cos (2 x)

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{e ^{3 x} x}{12} - \frac{5}{144} e^{3 x} + \frac{3}{40} e^{3 x} sin (2 x) + \frac{3}{40} e^{3 x} cos (2 x)

y^{^{''''}} + 2 y^{^{''}} + y = cos (2 x) + 5

y^{^{''}} + 16 y = cos (4 t), y (0) = 1, y (0) = 0

y^{^{''}} + 4 y^{^{'}} + 20 x = 0

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 5 y = x^{3}

y^{^{'''}} + 4 y^{^{''}} + 4 y^{^{'}} = 11 + 5 e^{2 x}