What is the solution for y^{'''}+y^{''}=9x^2 ?

The solution for y^{'''}+y^{''}=9x^2 is y=c_{1}e^{-x}+c_{2}+c_{3}x+(3x^4)/4-3x^3+9x^2

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y^{'''}+y^{''}=9x^2

y^{^{'''}} + y^{^{''}} = 9 x^{2}

y = c_{1} e^{- x} + c_{2} + c_{3} x + \frac{3 x ^{4}}{4} - 3 x^{3} + 9 x^{2}

Solution steps

y^{'''} (x) + y^{''} (x) = 9 x^{2}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} + c_{3} x + \frac{3 x ^{4}}{4} - 3 x^{3} + 9 x^{2}

y = c_{1} e^{- x} + c_{2} + c_{3} x + \frac{3 x ^{4}}{4} - 3 x^{3} + 9 x^{2}

y^{^{''}} + 2 y^{^{'}} - 3 y = 7 cos (3 x)

y^{^{''}} - 9 y^{^{'}} + 4 y = x e^{x}

y^{^{''}} + 4 y = sec^{2} (x)

y^{^{''}} + 2 y^{^{'}} - 3 y = 1, y (0) = e^{x} - \frac{1}{3}

y^{^{'''}} - 12 y^{^{''}} + 256 y = e^{8 x}

What is the solution for y^{'''}+y^{''}=9x^2 ?
The solution for y^{'''}+y^{''}=9x^2 is y=c_{1}e^{-x}+c_{2}+c_{3}x+(3x^4)/4-3x^3+9x^2