What is the solution for y^{''}+4y^'+5y=-10+e^{-x} ?

The solution for y^{''}+4y^'+5y=-10+e^{-x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))-2+1/2 e^{-x}

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y^{''}+4y^'+5y=-10+e^{-x}

y^{^{''}} + 4 y^{^{'}} + 5 y = - 10 + e^{- x}

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = - 10 + e^{- x}

Solve linear ODE:

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

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

y^{^{''}} - 9 y = e^{3 x}, y (0) = 1, y^{^{'}} (0) = - 5

y^{^{''}} + y^{^{'}} - 6 y = e^{- 5 x} (- 102 + 146 x - 70 x^{2})

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

y^{^{''}} + 2 y^{^{'}} = 2 x + 5 - 2 e^{- 2 x}

y^{^{''}} + 25 y = x + sin (x)

What is the solution for y^{''}+4y^'+5y=-10+e^{-x} ?
The solution for y^{''}+4y^'+5y=-10+e^{-x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))-2+1/2 e^{-x}