y^{'''}+64y^'=sec(8x)

解

y^{^{'''}} + 64 y^{^{'}} = sec (8 x)

y = c_{1} + c_{2} cos (8 x) + c_{3} sin (8 x) + \frac{1}{512} ln (tan (8 x) + sec (8 x)) - \frac{1}{64} x cos (8 x) + \frac{1}{512} ln (cos (8 x)) sin (8 x)

解答ステップ

y^{'''} (x) + 64 y^{'} (x) = sec (8 x)

線形 ODE を解く:

y = c_{1} + c_{2} cos (8 x) + c_{3} sin (8 x) + \frac{1}{512} ln (tan (8 x) + sec (8 x)) - \frac{1}{64} x cos (8 x) + \frac{1}{512} ln (cos (8 x)) sin (8 x)

y = c_{1} + c_{2} cos (8 x) + c_{3} sin (8 x) + \frac{1}{512} ln (tan (8 x) + sec (8 x)) - \frac{1}{64} x cos (8 x) + \frac{1}{512} ln (cos (8 x)) sin (8 x)