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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{1}{9} ln (ln (x)) + 3)

x^{3} y^{^{'}} = y

f (x) = (2 - 4 x^{2})^{5}, at x = 1

y^{^{'}} + 8 (tan (8 x)) y = 3 cos (8 x)

f (x) = x^{3} - 4 x^{2} - 6 x + 8

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