y^{''}+4y^'+6y=cos(x),y(0)=0,y^'(0)=2

解

y^{^{''}} + 4 y^{^{'}} + 6 y = cos (x), y (0) = 0, y^{^{'}} (0) = 2

y = e^{- 2 x} (- \frac{5}{41} cos (2 x) + \frac{34 2}{41} sin (2 x)) + \frac{4}{41} sin (x) + \frac{5}{41} cos (x)

解答ステップ

y^{''} (x) + 4 y^{'} (x) + 6 y = cos (x)

線形 ODE を解く:

y = e^{- 2 x} (- \frac{5}{41} cos (2 x) + \frac{34 2}{41} sin (2 x)) + \frac{4}{41} sin (x) + \frac{5}{41} cos (x)

y = e^{- 2 x} (- \frac{5}{41} cos (2 x) + \frac{34 2}{41} sin (2 x)) + \frac{4}{41} sin (x) + \frac{5}{41} cos (x)