y^{'''}+y^'=t^2+1

Lösung

y^{^{'''}} + y^{^{'}} = t^{2} + 1

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

Schritte zur Lösung

y^{'''} (t) + y^{'} (t) = t^{2} + 1

Solve linear ODE:

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

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

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

y^{^{''}} - 9 y^{^{'}} = 81 x^{2} + 7, y (0) = - 1, y^{^{'}} (0) = 2

y^{^{''}} - y^{^{'}} - 6 y = 14 sin (3)

y^{^{''}} + 2 y^{^{'}} + 5 y = 2 e^{(- t)} sin (2 t)

x^{^{''}} - 4 x^{^{'}} = 20 cos (5 t) - 4 sin (5 t) - 7