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}

\frac{\partial}{\partial x} (e^{y - x})

x \to 0 - lim (x + \frac{∣ x ∣}{x})

\frac{d y}{d x} + 3 y = \frac{3}{2}

\frac{d}{d x} (e^{ik x})

\frac{d}{d x} ((2 x - 3)^{4})

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}