y^{''}+2y^'+2y=((e^{-x}))/(cos(x))

Lösung

y^{^{''}} + 2 y^{^{'}} + 2 y = \frac{( e ^{- x} )}{cos ( x )}

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + ln (cos (x)) e^{- x} cos (x) + x e^{- x} sin (x)

Schritte zur Lösung

y^{''} (x) + 2 y^{'} (x) + 2 y = \frac{( e ^{- x} )}{cos ( x )}

Solve linear ODE:

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + ln (cos (x)) e^{- x} cos (x) + x e^{- x} sin (x)

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + ln (cos (x)) e^{- x} cos (x) + x e^{- x} sin (x)

\frac{d ^{2} y}{d x ^{2}} + 4 y = 2 sin (2 x)

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 \cdot e^{- t}

y^{^{''}} + 16 y = 5 cos (4 t), y (0) = 1, y^{^{'}} (0) = 0

y^{^{''''}} + 9 y^{^{''}} = 20 e^{- t}

y^{^{''''}} - y = 9 t + cos (t)