y^{''''}+25y^{''}=sin(5t)+te^{6t}+4

解

y^{^{''''}} + 25 y^{^{''}} = sin (5 t) + t e^{6 t} + 4

y = c_{1} + c_{2} t + c_{3} cos (5 t) + c_{4} sin (5 t) + \frac{t cos ( 5 t )}{250} + \frac{e ^{6 t} t}{2196} - \frac{97}{401868} e^{6 t} + \frac{2 t ^{2}}{25}

解答ステップ

y^{(4)} (t) + 25 y^{''} (t) = sin (5 t) + t e^{6 t} + 4

線形 ODE を解く:

y = c_{1} + c_{2} t + c_{3} cos (5 t) + c_{4} sin (5 t) + \frac{t cos ( 5 t )}{250} + \frac{e ^{6 t} t}{2196} - \frac{97}{401868} e^{6 t} + \frac{2 t ^{2}}{25}

y = c_{1} + c_{2} t + c_{3} cos (5 t) + c_{4} sin (5 t) + \frac{t cos ( 5 t )}{250} + \frac{e ^{6 t} t}{2196} - \frac{97}{401868} e^{6 t} + \frac{2 t ^{2}}{25}