y^{'''}-5y^{''}+9y=tcos(2t)

解

y^{^{'''}} - 5 y^{^{''}} + 9 y = t cos (2 t)

y = c_{1} e^{- 1.20440 \dots t} + c_{2} e^{1.63555 \dots t} + c_{3} e^{4.56884 \dots t} - \frac{8 t sin ( 2 t )}{905} + \frac{29 t cos ( 2 t )}{905} - \frac{21108}{819025} sin (2 t) + \frac{44}{819025} cos (2 t)

解答ステップ

y^{'''} (t) - 5 y^{''} (t) + 9 y = t cos (2 t)

線形 ODE を解く:

y = c_{1} e^{- 1.20440 \dots t} + c_{2} e^{1.63555 \dots t} + c_{3} e^{4.56884 \dots t} - \frac{8 t sin ( 2 t )}{905} + \frac{29 t cos ( 2 t )}{905} - \frac{21108}{819025} sin (2 t) + \frac{44}{819025} cos (2 t)

y = c_{1} e^{- 1.20440 \dots t} + c_{2} e^{1.63555 \dots t} + c_{3} e^{4.56884 \dots t} - \frac{8 t sin ( 2 t )}{905} + \frac{29 t cos ( 2 t )}{905} - \frac{21108}{819025} sin (2 t) + \frac{44}{819025} cos (2 t)