y^{''}-6y^'+25y=sin(t/2)+e^{-t}t

Lösung

y^{^{''}} - 6 y^{^{'}} + 25 y = sin (\frac{t}{2}) + e^{- t} t

y = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + \frac{44}{1105} sin (\frac{t}{2}) + \frac{16}{3315} cos (\frac{t}{2}) + \frac{e ^{- t} t}{32} + \frac{1}{128} e^{- t}

Schritte zur Lösung

y^{''} (t) - 6 y^{'} (t) + 25 y = sin (\frac{t}{2}) + e^{- t} t

Solve linear ODE:

y = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + \frac{44}{1105} sin (\frac{t}{2}) + \frac{16}{3315} cos (\frac{t}{2}) + \frac{e ^{- t} t}{32} + \frac{1}{128} e^{- t}

y = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + \frac{44}{1105} sin (\frac{t}{2}) + \frac{16}{3315} cos (\frac{t}{2}) + \frac{e ^{- t} t}{32} + \frac{1}{128} e^{- t}

y^{^{''''}} + 2 y^{^{'''}} + 2 y^{^{''}} = 3 e^{x} + 2 x e^{- x}

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

(D^{2} + 4 D + 5) y = 50 x + 13 e^{x}

(d^{2} - 4) y (d, t) = 5 e^{3 t}, y (0) = 1

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