y^{''''}+4y^{''}=sin(2t)+te^{3t}+4

Lösung

y^{^{''''}} + 4 y^{^{''}} = sin (2 t) + t e^{3 t} + 4

y = c_{1} + c_{2} t + c_{3} cos (2 t) + c_{4} sin (2 t) + \frac{t cos ( 2 t )}{16} + \frac{e ^{3 t} t}{117} - \frac{44}{4563} e^{3 t} + \frac{t ^{2}}{2}

Schritte zur Lösung

y^{(4)} (t) + 4 y^{''} (t) = sin (2 t) + t e^{3 t} + 4

Solve linear ODE:

y = c_{1} + c_{2} t + c_{3} cos (2 t) + c_{4} sin (2 t) + \frac{t cos ( 2 t )}{16} + \frac{e ^{3 t} t}{117} - \frac{44}{4563} e^{3 t} + \frac{t ^{2}}{2}

y = c_{1} + c_{2} t + c_{3} cos (2 t) + c_{4} sin (2 t) + \frac{t cos ( 2 t )}{16} + \frac{e ^{3 t} t}{117} - \frac{44}{4563} e^{3 t} + \frac{t ^{2}}{2}

y^{^{''}} - y^{^{'}} - 12 = 0

y^{^{'''}} + y = sin (x)

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

y^{^{''}} - 3 + y^{^{'}} + 2 y = 0

y^{^{''}} - 4 y = 4 x