y^{''}+4pi^2y=cos^3(2pix),y(0)=1

解

y^{^{''}} + 4 π^{2} y = cos^{3} (2 π x), y (0) = 1

y = (1 - \frac{1}{16 π ^{2}}) 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)))

解答ステップ

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

線形 ODE を解く:

y = (1 - \frac{1}{16 π ^{2}}) 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 = (1 - \frac{1}{16 π ^{2}}) 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)))