16y^{''''}+40y^{''}+25y=0

解

16 y^{^{''''}} + 40 y^{^{''}} + 25 y = 0

y = c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2}) + t (c_{3} cos (\frac{5 t}{2}) + c_{4} sin (\frac{5 t}{2}))

解答ステップ

16 y^{(4)} (t) + 40 y^{''} (t) + 25 y = 0

線形 ODE を解く:

y = c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2}) + t (c_{3} cos (\frac{5 t}{2}) + c_{4} sin (\frac{5 t}{2}))

y = c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2}) + t (c_{3} cos (\frac{5 t}{2}) + c_{4} sin (\frac{5 t}{2}))