What is the solution for y^{''}+3y^'+2y=e^{-2x}+x ?

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

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

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

y = c_{1} e^{- x} + c_{2} e^{- 2 x} - e^{- 2 x} x + \frac{x}{2} - \frac{3}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 2 x} - e^{- 2 x} x + \frac{x}{2} - \frac{3}{4}

y = c_{1} e^{- x} + c_{2} e^{- 2 x} - e^{- 2 x} x + \frac{x}{2} - \frac{3}{4}

2 x^{^{''}} + 4 x^{^{'}} + 128 x = cos (3 t)

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

3 x^{^{''}} + 48 x = 180 e^{- 4 t} cos (4 t)

y^{^{''}} + 20 y^{^{'}} + 100 y = 240 cos (10 x)

y^{^{''}} + 3 y^{^{'}} - 4 y = cos (2 x), y (0) = 1

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