y^{''}-0.2y^'+0.01y=0,y(0)=14,y^'(0)=b

解

y^{^{''}} - 0.2 y^{^{'}} + 0.01 y = 0, y (0) = 14, y^{^{'}} (0) = b

y = 14 e^{\frac{t}{10}} + (b - \frac{7}{5}) e^{\frac{t}{10}} t

解答ステップ

y^{''} (t) - 0.2 y^{'} (t) + 0.01 y = 0

線形 ODE を解く:

y = 14 e^{\frac{t}{10}} + (b - \frac{7}{5}) e^{\frac{t}{10}} t

y = 14 e^{\frac{t}{10}} + (b - \frac{7}{5}) e^{\frac{t}{10}} t