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

Lösung

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}

Schritte zur Lösung

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

Solve linear ODE:

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}

y^{^{''}} + y^{^{'}} = e^{5 x}

y^{^{''}} - 16 y^{^{'}} + 64 y = 6 e^{8 x}

y^{^{''}} - 4 y^{^{'}} + 3 y = sin (e^{- x})

y^{^{''}} + 2 y^{^{'}} + y = x ln (x)

(D^{3} - 3 D - 2) y = x^{2}