y^{''}-y^'+2y=cos(x),y(0)=1,y^'(0)=0

解

y^{^{''}} - y^{^{'}} + 2 y = cos (x), y (0) = 1, y^{^{'}} (0) = 0

y = e^{\frac{x}{2}} (\frac{1}{2} cos (\frac{7 x}{2}) + \frac{1}{2 7} sin (\frac{7 x}{2})) - \frac{1}{2} sin (x) + \frac{1}{2} cos (x)

解答ステップ

y^{''} (x) - y^{'} (x) + 2 y = cos (x)

線形 ODE を解く:

y = e^{\frac{x}{2}} (\frac{1}{2} cos (\frac{7 x}{2}) + \frac{1}{2 7} sin (\frac{7 x}{2})) - \frac{1}{2} sin (x) + \frac{1}{2} cos (x)

y = e^{\frac{x}{2}} (\frac{1}{2} cos (\frac{7 x}{2}) + \frac{1}{2 7} sin (\frac{7 x}{2})) - \frac{1}{2} sin (x) + \frac{1}{2} cos (x)