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

解

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

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{1}{8} ln (tan (2 x) + sec (2 x)) - \frac{1}{4} x cos (2 x) + \frac{1}{8} ln (cos (2 x)) sin (2 x)

解答ステップ

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

線形 ODE を解く:

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{1}{8} ln (tan (2 x) + sec (2 x)) - \frac{1}{4} x cos (2 x) + \frac{1}{8} ln (cos (2 x)) sin (2 x)

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{1}{8} ln (tan (2 x) + sec (2 x)) - \frac{1}{4} x cos (2 x) + \frac{1}{8} ln (cos (2 x)) sin (2 x)