What is the solution for y^{''}-6y^'+8y^'=x^2-1 ?

The solution for y^{''}-6y^'+8y^'=x^2-1 is y=c_{1}+c_{2}e^{-2x}+(x^3)/6-(x^2)/4-x/4

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y^{''}-6y^'+8y^'=x^2-1

y^{^{''}} - 6 y^{^{'}} + 8 y^{^{'}} = x^{2} - 1

y = c_{1} + c_{2} e^{- 2 x} + \frac{x ^{3}}{6} - \frac{x ^{2}}{4} - \frac{x}{4}

Solution steps

y^{''} (x) - 6 y^{'} (x) + 8 y^{'} (x) = x^{2} - 1

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} + \frac{x ^{3}}{6} - \frac{x ^{2}}{4} - \frac{x}{4}

y = c_{1} + c_{2} e^{- 2 x} + \frac{x ^{3}}{6} - \frac{x ^{2}}{4} - \frac{x}{4}

What is the solution for y^{''}-6y^'+8y^'=x^2-1 ?
The solution for y^{''}-6y^'+8y^'=x^2-1 is y=c_{1}+c_{2}e^{-2x}+(x^3)/6-(x^2)/4-x/4