x^{''}+4x=sin^2(2t)

解答

x^{^{''}} + 4 x = sin^{2} (2 t)

x = c_{1} cos (2 t) + c_{2} sin (2 t) + (cos^{4} (t) - \frac{2}{3} cos^{6} (t)) cos (2 t) + \frac{1}{24} (sin (6 t) - 3 sin (2 t) cos (4 t)) sin (2 t)

求解步骤

x^{''} (t) + 4 x = sin^{2} (2 t)

求解线性方程:

x = c_{1} cos (2 t) + c_{2} sin (2 t) + (cos^{4} (t) - \frac{2}{3} cos^{6} (t)) cos (2 t) + \frac{1}{24} (sin (6 t) - 3 sin (2 t) cos (4 t)) sin (2 t)

x = c_{1} cos (2 t) + c_{2} sin (2 t) + (cos^{4} (t) - \frac{2}{3} cos^{6} (t)) cos (2 t) + \frac{1}{24} (sin (6 t) - 3 sin (2 t) cos (4 t)) sin (2 t)