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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + 8 y^{^{'}} + 16 y = \frac{3 e ^{- 4 x}}{x}

(d^{3} - 3 d^{2} - 10 d + 24) y = 2 x^{2} + 5

y^{^{''}} - y^{^{'}} + 2 = 0

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

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