(d^4y)/(dt^4)-4y^{''}=t^2+e^t

Lösung

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

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + c_{3} + c_{4} t - \frac{t ^{4}}{48} - \frac{t ^{2}}{16} - \frac{1}{3} e^{t}

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + c_{3} + c_{4} t - \frac{t ^{4}}{48} - \frac{t ^{2}}{16} - \frac{1}{3} e^{t}

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + c_{3} + c_{4} t - \frac{t ^{4}}{48} - \frac{t ^{2}}{16} - \frac{1}{3} e^{t}

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