y^{''}+4pi^2y=cos^3(2pix)

解答

y^{^{''}} + 4 π^{2} y = cos^{3} (2 π x)

y = c_{1} cos (2 π x) + c_{2} sin (2 π x) + \frac{1}{16 π ^{2}} cos^{5} (2 π x) + \frac{1}{4 π ^{2}} sin (2 π x) (cos^{3} (2 π x) sin (2 π x) + \frac{3}{8} (2 π x - \frac{1}{4} sin (8 π x)))

求解步骤

y^{''} (x) + 4 π^{2} y = cos^{3} (2 π x)

求解线性方程:

y = c_{1} cos (2 π x) + c_{2} sin (2 π x) + \frac{1}{16 π ^{2}} cos^{5} (2 π x) + \frac{1}{4 π ^{2}} sin (2 π x) (cos^{3} (2 π x) sin (2 π x) + \frac{3}{8} (2 π x - \frac{1}{4} sin (8 π x)))

y = c_{1} cos (2 π x) + c_{2} sin (2 π x) + \frac{1}{16 π ^{2}} cos^{5} (2 π x) + \frac{1}{4 π ^{2}} sin (2 π x) (cos^{3} (2 π x) sin (2 π x) + \frac{3}{8} (2 π x - \frac{1}{4} sin (8 π x)))

\int 2 x I n x^{2} d x

\frac{d}{d x} (\frac{8 ^{x}}{7 ^{x}})

\int \frac{x ^{2}}{9 + 4 x ^{2}} d x

\int 3 x^{2} sin (x^{3} - 1) d x

d er i v a t i v e - 4 t (3 t^{5} - 1)^{4}