What is the solution for y^{''}+4y^'+8y=x+e^x ?

The solution for y^{''}+4y^'+8y=x+e^x is y=e^{-2x}(c_{1}cos(2x)+c_{2}sin(2x))+x/8-1/16+1/13 e^x

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y^{''}+4y^'+8y=x+e^x

y^{^{''}} + 4 y^{^{'}} + 8 y = x + e^{x}

y = e^{- 2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{x}{8} - \frac{1}{16} + \frac{1}{13} e^{x}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 8 y = x + e^{x}

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{x}{8} - \frac{1}{16} + \frac{1}{13} e^{x}

y = e^{- 2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{x}{8} - \frac{1}{16} + \frac{1}{13} e^{x}

4 y^{^{''}} + 11 y^{^{'}} - 3 y = 2 x e^{- 3 x}

y^{^{''}} + 4 y = 2 e^{2 t} + 4 cos (3 t)

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- x} (3 - 4 x)

y^{^{''}} - 16 y^{^{'}} = 2 e^{4 x}

y^{^{'''}} + 4 y^{^{'}} = t^{2} + 2 t

What is the solution for y^{''}+4y^'+8y=x+e^x ?
The solution for y^{''}+4y^'+8y=x+e^x is y=e^{-2x}(c_{1}cos(2x)+c_{2}sin(2x))+x/8-1/16+1/13 e^x