What is the solution for y^{'''}-y^'=1-x^2 ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 e^{(3 t)}

y^{^{''}} + 4 y = 3 x

y^{^{''}} + 2 a y^{^{'}} + b^{2} y = sin (c t)

x^{^{''}} + 10 x^{^{'}} + 16 x = 2581 cos (5 t)

(D^{4} + 10 D^{2} + 9) y = sin (3 x) + cos (3 x)

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