(d^2+9)y=30sin^3(x)

解答

(d^{2} + 9) y = 30 sin^{3} (x)

y = c_{1} cos (3 x) + c_{2} sin (3 x) - 10 cos (3 x) (- sin (3 x) cos (x) + \frac{1}{3} cos^{3} (x) sin (3 x) - \frac{1}{8} x + \frac{13}{32} sin (4 x) + \frac{23}{32} sin (2 x) - \frac{1}{8} cos^{2} (x) sin (4 x) - \frac{1}{4} cos^{2} (x) sin (2 x) + \frac{1}{96} sin (6 x)) + sin (3 x) \cdot \int 10 sin^{3} (x) cos (3 x) d x

求解步骤

(d^{2} + 9) y = 30 sin^{3} (x)

求解线性方程:

y = c_{1} cos (3 x) + c_{2} sin (3 x) - 10 cos (3 x) (- sin (3 x) cos (x) + \frac{1}{3} cos^{3} (x) sin (3 x) - \frac{1}{8} x + \frac{13}{32} sin (4 x) + \frac{23}{32} sin (2 x) - \frac{1}{8} cos^{2} (x) sin (4 x) - \frac{1}{4} cos^{2} (x) sin (2 x) + \frac{1}{96} sin (6 x)) + sin (3 x) \cdot \int 10 sin^{3} (x) cos (3 x) d x

y = c_{1} cos (3 x) + c_{2} sin (3 x) - 10 cos (3 x) (- sin (3 x) cos (x) + \frac{1}{3} cos^{3} (x) sin (3 x) - \frac{1}{8} x + \frac{13}{32} sin (4 x) + \frac{23}{32} sin (2 x) - \frac{1}{8} cos^{2} (x) sin (4 x) - \frac{1}{4} cos^{2} (x) sin (2 x) + \frac{1}{96} sin (6 x)) + sin (3 x) \cdot \int 10 sin^{3} (x) cos (3 x) d x

y^{^{''''}} - y = 11 t + cos (t)

y^{^{''}} - 4 y^{^{'}} + 4 y = 10 e^{2 x}

2 \frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 3 = 0

(d^{3} - 3 d^{2} + 4 d - 2) y = x^{3}

y^{^{''}} = 2 y + tan^{3} (x), y (0) = \frac{1}{2} tan (x)