y^{''}+2y^'+2y=e^{-x}(cos(x)+1)

Lösung

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{- x} (cos (x) + 1)

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

Schritte zur Lösung

y^{''} (x) + 2 y^{'} (x) + 2 y = e^{- x} (cos (x) + 1)

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 8 y = e^{x} cos (2 x)

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

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

y^{^{'''}} - 25 y^{^{'}} = t e^{- 5 t} + 6 cos (5 t)

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{t}}{t ^{2} + 1}