y^{'''}+25y^'=tan(5x)

Lösung

y^{^{'''}} + 25 y^{^{'}} = tan (5 x)

y = c_{1} + c_{2} cos (5 x) + c_{3} sin (5 x) + \frac{- ln ( cos ( 5 x ) ) + cos ^{2} ( 5 x ) - sin ( 5 x ) ( ln ( tan ( 5 x ) + sec ( 5 x ) ) - sin ( 5 x ) )}{125}

Schritte zur Lösung

y^{'''} (x) + 25 y^{'} (x) = tan (5 x)

Solve linear ODE:

y = c_{1} + c_{2} cos (5 x) + c_{3} sin (5 x) + \frac{- ln ( cos ( 5 x ) ) + cos ^{2} ( 5 x ) - sin ( 5 x ) ( ln ( tan ( 5 x ) + sec ( 5 x ) ) - sin ( 5 x ) )}{125}

y = c_{1} + c_{2} cos (5 x) + c_{3} sin (5 x) + \frac{- ln ( cos ( 5 x ) ) + cos ^{2} ( 5 x ) - sin ( 5 x ) ( ln ( tan ( 5 x ) + sec ( 5 x ) ) - sin ( 5 x ) )}{125}

y^{^{''}} - 2 y^{^{'}} + 5 y = - 8

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

\frac{d ^{2} x}{d t ^{2}} + 81 x = 2 sin (9 t)

(D^{2} - 1) y = e^{- x} (2 sin (x) + 4 cos (x))

\frac{d ^{2} y}{d x ^{2}} - y = x \cdot e^{2 x}