y^{'''}+4y^'=t^2+2t

Lösung

y^{^{'''}} + 4 y^{^{'}} = t^{2} + 2 t

y = c_{1} + c_{2} cos (2 t) + c_{3} sin (2 t) + \frac{t ^{3}}{12} + \frac{t ^{2}}{4} - \frac{t}{8}

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} + c_{2} cos (2 t) + c_{3} sin (2 t) + \frac{t ^{3}}{12} + \frac{t ^{2}}{4} - \frac{t}{8}

y = c_{1} + c_{2} cos (2 t) + c_{3} sin (2 t) + \frac{t ^{3}}{12} + \frac{t ^{2}}{4} - \frac{t}{8}

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

4 y^{^{''}} + 8 y^{^{'}} + 5 y = e^{- x} sec (\frac{x}{2})

y^{^{''}} + 2 y^{^{'}} = 24 x + e^{- 2 x}

x^{^{''}} + 16 x = sin (t)

\frac{d ^{2} y}{d t ^{2}} + 8 \frac{d y}{d t} - 20 y = 5 e^{- 3 t}