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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 16 y = 32 + 45 e^{x}

d (x) = x^{2} - 8 x + 15

y^{^{''}} + 3 y^{^{'}} - 4 y = 36 e^{5 t}

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 3 + 7 x

y^{^{''}} - y = \frac{e ^{2 x} - e ^{- 2 x}}{2}

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