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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(d^{2} + 9) y = 7 e^{3 x} + 3 x^{2} + 4

y^{^{''}} - 2 y^{^{'}} + y = (3 x + 1) e^{x}

(d^{2} - 13 d + 12) \cdot y = x

(d - 1)^{2} \cdot y = e^{x} \cdot sec^{2} (x) tan (x)

y^{^{''}} + y^{^{'}} - 2 y = 20 sin (2 x)

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