10y^{''}-50y^'+65y=0,y(0)=1,y^'(0)=1

解

10 y^{^{''}} - 50 y^{^{'}} + 65 y = 0, y (0) = 1, y^{^{'}} (0) = 1

y = e^{\frac{5 t}{2}} (cos (\frac{t}{2}) - 3 sin (\frac{t}{2}))

解答ステップ

10 y^{''} (t) - 50 y^{'} (t) + 65 y = 0

線形 ODE を解く:

y = e^{\frac{5 t}{2}} (cos (\frac{t}{2}) - 3 sin (\frac{t}{2}))

y = e^{\frac{5 t}{2}} (cos (\frac{t}{2}) - 3 sin (\frac{t}{2}))