x^{'''}+3x^{''}+x^'+3x=t^2

解

x^{^{'''}} + 3 x^{^{''}} + x^{^{'}} + 3 x = t^{2}

x = c_{1} e^{- 3 t} + c_{2} cos (t) + c_{3} sin (t) + \frac{t ^{2}}{3} - \frac{2 t}{9} - \frac{16}{27}

解答ステップ

x^{'''} (t) + 3 x^{''} (t) + x^{'} (t) + 3 x = t^{2}

線形 ODE を解く:

x = c_{1} e^{- 3 t} + c_{2} cos (t) + c_{3} sin (t) + \frac{t ^{2}}{3} - \frac{2 t}{9} - \frac{16}{27}

x = c_{1} e^{- 3 t} + c_{2} cos (t) + c_{3} sin (t) + \frac{t ^{2}}{3} - \frac{2 t}{9} - \frac{16}{27}