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

解

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + (- \frac{1}{4} x sin^{4} (x) - \frac{1}{16} sin^{3} (x) cos (x) + \frac{3}{32} x - \frac{3}{64} sin (2 x)) cos (2 x) + \frac{1}{128} (- 8 x^{2} + 32 x sin^{2} (x) sin (2 x) + 4 x sin (4 x) - 16 sin^{4} (x) + cos (4 x)) sin (2 x)

解答ステップ

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

線形 ODE を解く:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + (- \frac{1}{4} x sin^{4} (x) - \frac{1}{16} sin^{3} (x) cos (x) + \frac{3}{32} x - \frac{3}{64} sin (2 x)) cos (2 x) + \frac{1}{128} (- 8 x^{2} + 32 x sin^{2} (x) sin (2 x) + 4 x sin (4 x) - 16 sin^{4} (x) + cos (4 x)) sin (2 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + (- \frac{1}{4} x sin^{4} (x) - \frac{1}{16} sin^{3} (x) cos (x) + \frac{3}{32} x - \frac{3}{64} sin (2 x)) cos (2 x) + \frac{1}{128} (- 8 x^{2} + 32 x sin^{2} (x) sin (2 x) + 4 x sin (4 x) - 16 sin^{4} (x) + cos (4 x)) sin (2 x)