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

The solution for y^{'''}+y^{''}=x^2 is y=c_{1}e^{-x}+c_{2}+c_{3}x+(x^4}{12}-\frac{x^3)/3+x^2

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

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

y = c_{1} e^{- x} + c_{2} + c_{3} x + \frac{x ^{4}}{12} - \frac{x ^{3}}{3} + x^{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} + c_{3} x + \frac{x ^{4}}{12} - \frac{x ^{3}}{3} + x^{2}

y = c_{1} e^{- x} + c_{2} + c_{3} x + \frac{x ^{4}}{12} - \frac{x ^{3}}{3} + x^{2}

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

y^{^{'''}} - 4 y^{^{''}} + 4 y^{^{'}} = x^{2} + 4 e^{2 x}

y^{^{''}} - y = 2 x^{4} + 5 x^{2} - 10 x

y^{^{''}} + 2 y^{^{'}} + 5 y = 4 e^{(- x)} - x cos (2 x)

(D^{2} + 1) y = sin (x) sin (2 x)

What is the solution for y^{'''}+y^{''}=x^2 ?
The solution for y^{'''}+y^{''}=x^2 is y=c_{1}e^{-x}+c_{2}+c_{3}x+(x^4}{12}-\frac{x^3)/3+x^2