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}

y^{^{''}} + 4 y = 4

y^{^{''}} + y^{^{'}} - 12 y = 2 cos (t), y (0) = 1, y^{^{'}} (0) = 1

y^{^{'''}} + 8 y = 2 x - 5 + 8 e^{- 2 x}

y^{^{''}} - y = cosh (t)

d (x) = - 6 x + 4200

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}