2x^{''}+3x^'+5x=2t

解

2 x^{^{''}} + 3 x^{^{'}} + 5 x = 2 t

x = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{31 t}{4}) + c_{2} sin (\frac{31 t}{4})) + \frac{2 t}{5} - \frac{6}{25}

解答ステップ

2 x^{''} (t) + 3 x^{'} (t) + 5 x = 2 t

線形 ODE を解く:

x = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{31 t}{4}) + c_{2} sin (\frac{31 t}{4})) + \frac{2 t}{5} - \frac{6}{25}

x = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{31 t}{4}) + c_{2} sin (\frac{31 t}{4})) + \frac{2 t}{5} - \frac{6}{25}