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

解

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

y = (\frac{1}{3} + \frac{4}{5}) e^{x} + \frac{1}{6} e^{- 2 x} - \frac{x}{2} - \frac{1}{4} - \frac{3}{20} sin (2 x) - \frac{1}{20} cos (2 x)

解答ステップ

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

線形 ODE を解く:

y = (\frac{1}{3} + \frac{4}{5}) e^{x} + \frac{1}{6} e^{- 2 x} - \frac{x}{2} - \frac{1}{4} - \frac{3}{20} sin (2 x) - \frac{1}{20} cos (2 x)

y = (\frac{1}{3} + \frac{4}{5}) e^{x} + \frac{1}{6} e^{- 2 x} - \frac{x}{2} - \frac{1}{4} - \frac{3}{20} sin (2 x) - \frac{1}{20} cos (2 x)