What is the solution for y^{''}-y^'=2x^2-3 ?

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

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y^{''}-y^'=2x^2-3

y^{^{''}} - y^{^{'}} = 2 x^{2} - 3

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - 8 y^{^{'}} + 80 y = - 3 sin (3 t)

y^{^{''}} - 8 y^{^{'}} + 16 = (1 - x) e^{4 x}

y^{^{'''}} + 4 y^{^{''}} - 5 y = x e^{x} + 14 x

y^{^{''}} - 2 y^{^{'}} + y = - 5 e^{x} + 8 x^{2}

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