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

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

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

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

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{2} e^{- x} - x + \frac{4}{5}

Solution steps

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

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{2} e^{- x} - x + \frac{4}{5}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{2} e^{- x} - x + \frac{4}{5}

f (x) = 3 x^{2} - 10 x

π x^{3}

y = e^{2 x} tan (2 x)

y = 3 x^{2} - x + 1

x \frac{d y}{d x} + 3 (y + x^{2}) = \frac{sin ( x )}{x}

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