y^{'''}+49y^'=tan(7x)

Lösung

y^{^{'''}} + 49 y^{^{'}} = tan (7 x)

y = c_{1} + c_{2} cos (7 x) + c_{3} sin (7 x) + \frac{- ln ( cos ( 7 x ) ) + cos ^{2} ( 7 x ) - sin ( 7 x ) ( ln ( tan ( 7 x ) + sec ( 7 x ) ) - sin ( 7 x ) )}{343}

Schritte zur Lösung

y^{'''} (x) + 49 y^{'} (x) = tan (7 x)

Solve linear ODE:

y = c_{1} + c_{2} cos (7 x) + c_{3} sin (7 x) + \frac{- ln ( cos ( 7 x ) ) + cos ^{2} ( 7 x ) - sin ( 7 x ) ( ln ( tan ( 7 x ) + sec ( 7 x ) ) - sin ( 7 x ) )}{343}

y = c_{1} + c_{2} cos (7 x) + c_{3} sin (7 x) + \frac{- ln ( cos ( 7 x ) ) + cos ^{2} ( 7 x ) - sin ( 7 x ) ( ln ( tan ( 7 x ) + sec ( 7 x ) ) - sin ( 7 x ) )}{343}

y^{^{''}} + a y^{^{'}} = e^{3 t}

4 y^{^{''}} - 11 y^{^{'}} - 3 y = 5 e^{3 x}

l in e a r y^{^{''}} - 3 y^{^{'}} + 2 y = e^{x}

y^{^{''}} - y^{^{'}} + y = (2 e^{3 t} + 5 t)^{7}

\frac{d ^{2} y}{d x ^{2}} - 9 \frac{d y}{d x} + 5 y = x e^{x}