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

解

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

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

解答ステップ

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

線形 ODE を解く:

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

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