y^{''}+y^'+0.25y=e^x(sin(3x)-cos(3x))

Lösung

y^{^{''}} + y^{^{'}} + 0.25 y = e^{x} (sin (3 x) - cos (3 x))

y = c_{1} e^{- \frac{x}{2}} + c_{2} e^{- \frac{x}{2}} x - 0.12444 \dots e^{x} sin (3 x) - 0.01777 \dots e^{x} cos (3 x)

Schritte zur Lösung

y^{''} (x) + y^{'} (x) + 0.25 y = e^{x} (sin (3 x) - cos (3 x))

Solve linear ODE:

y = c_{1} e^{- \frac{x}{2}} + c_{2} e^{- \frac{x}{2}} x - 0.12444 \dots e^{x} sin (3 x) - 0.01777 \dots e^{x} cos (3 x)

y = c_{1} e^{- \frac{x}{2}} + c_{2} e^{- \frac{x}{2}} x - 0.12444 \dots e^{x} sin (3 x) - 0.01777 \dots e^{x} cos (3 x)

y^{^{''}} + 4 y^{^{'}} = t^{2} + 6 e^{t}, y (0) = 0, y^{^{'}} (0) = 9

y^{^{''}} + 4 y^{^{'}} = t^{2} + 6 e^{t}, y (0) = 0, y^{^{'}} (0) = 5

y^{^{''}} + y = t^{2} sin (t)

y^{^{'''}} - 8 y^{^{''}} + 5 y^{^{'}} + y = 5 t

y^{^{''}} - 2 y^{^{'}} + y = x (e^{x}) + 4