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

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

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

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

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

Solution steps

y^{(4)} (x) - y^{''} (x) = 3 x^{2} - 2 x

Solve linear ODE:

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

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

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

\frac{1}{6} y^{^{''}} + 6 y = tan (6 x) - \frac{1}{3} e^{3 x}

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

y^{^{'''}} - y^{^{'}} = 10 cos (2 x)

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

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