y^{''}+4y=sin^2(x)

解

y^{^{''}} + 4 y = sin^{2} (x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{4} sin^{4} (x) cos (2 x) + \frac{1}{2} sin (2 x) (\frac{1}{2} sin^{2} (x) sin (2 x) - \frac{1}{4} (x - \frac{1}{4} sin (4 x)))

解答ステップ

y^{''} (x) + 4 y = sin^{2} (x)

線形 ODE を解く:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{4} sin^{4} (x) cos (2 x) + \frac{1}{2} sin (2 x) (\frac{1}{2} sin^{2} (x) sin (2 x) - \frac{1}{4} (x - \frac{1}{4} sin (4 x)))

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{4} sin^{4} (x) cos (2 x) + \frac{1}{2} sin (2 x) (\frac{1}{2} sin^{2} (x) sin (2 x) - \frac{1}{4} (x - \frac{1}{4} sin (4 x)))