y^{'''}+289y^'=tan(17x)

Lösung

y^{^{'''}} + 289 y^{^{'}} = tan (17 x)

y = c_{1} + c_{2} cos (17 x) + c_{3} sin (17 x) + \frac{- ln ( cos ( 17 x ) ) + cos ^{2} ( 17 x ) - sin ( 17 x ) ( ln ( tan ( 17 x ) + sec ( 17 x ) ) - sin ( 17 x ) )}{4913}

Schritte zur Lösung

y^{'''} (x) + 289 y^{'} (x) = tan (17 x)

Solve linear ODE:

y = c_{1} + c_{2} cos (17 x) + c_{3} sin (17 x) + \frac{- ln ( cos ( 17 x ) ) + cos ^{2} ( 17 x ) - sin ( 17 x ) ( ln ( tan ( 17 x ) + sec ( 17 x ) ) - sin ( 17 x ) )}{4913}

y = c_{1} + c_{2} cos (17 x) + c_{3} sin (17 x) + \frac{- ln ( cos ( 17 x ) ) + cos ^{2} ( 17 x ) - sin ( 17 x ) ( ln ( tan ( 17 x ) + sec ( 17 x ) ) - sin ( 17 x ) )}{4913}

y^{^{''}} - 3 y^{^{'}} - 4 y = t^{2}

y^{^{''}} - 2 y^{^{'}} + y = 2 e^{t}, y (0) = 1, y^{^{'}} (0) = 1

x^{^{''}} - x^{^{'}} - 6 x = 5 t

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e^{- t}

y^{^{''}} - y^{^{'}} + 4 y = x^{3} e^{2 x} + x e^{2 x}