What is the solution for y^{''}+y^'-2y=1+x ?

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

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y^{''}+y^'-2y=1+x

y^{^{''}} + y^{^{'}} - 2 y = 1 + x

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

Solution steps

y^{''} (x) + y^{'} (x) - 2 y = 1 + x

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 4 y = sin (4 x)

y^{^{''}} + 6 y^{^{'}} + 9 y = e^{3 x} ln (x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 6 e^{x} - 2 x e^{2 x} + 2

y^{^{''}} + 3 y^{^{'}} - 6 y = x e^{2 x}

y^{^{''}} + 2 y^{^{'}} = x + 1

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