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

The solution for y^{''}+4y^'+5y=-10x+e^{-x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))-2x+8/5+1/2 e^{-x}

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

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

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) - 2 x + \frac{8}{5} + \frac{1}{2} e^{- x}

Solution steps

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

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) - 2 x + \frac{8}{5} + \frac{1}{2} e^{- x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) - 2 x + \frac{8}{5} + \frac{1}{2} e^{- x}

t e^{2 t} sin (6 t)

\frac{d}{d x} (e^{x} cos (x) - x^{3} sin (x) + π)

y^{^{'}} = \frac{y}{50} 5, y (0) = 30

(7, 6), (- 7, - 7)

\frac{d y}{d x} = \frac{y}{x} + (\frac{y}{x})^{2} - 1

What is the solution for y^{''}+4y^'+5y=-10x+e^{-x} ?
The solution for y^{''}+4y^'+5y=-10x+e^{-x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))-2x+8/5+1/2 e^{-x}