What is the solution for y^{''}+4y^'+5y=-5x+3e^{(-x)} ?

The solution for y^{''}+4y^'+5y=-5x+3e^{(-x)} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))-x+4/5+3/2 e^{-x}

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y^{''}+4y^'+5y=-5x+3e^{(-x)}

y^{^{''}} + 4 y^{^{'}} + 5 y = - 5 x + 3 e^{(- x)}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) - x + \frac{4}{5} + \frac{3}{2} e^{- x}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = - 5 x + 3 e^{(- x)}

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) - x + \frac{4}{5} + \frac{3}{2} e^{- x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) - x + \frac{4}{5} + \frac{3}{2} e^{- x}

(d^{2} - 4 d + 4) y = 5 sin^{3} (x)

y^{^{''}} - 9 y = (x^{2} - 2) sin (3 x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 11 e^{3} x, y^{1} (0) = e x

y^{^{''}} + 4 y = 8 x^{2} + 7 x

y^{^{''}} - 4 y^{^{'}} = x

What is the solution for y^{''}+4y^'+5y=-5x+3e^{(-x)} ?
The solution for y^{''}+4y^'+5y=-5x+3e^{(-x)} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))-x+4/5+3/2 e^{-x}