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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

5 y^{^{''}} + 7 y^{^{'}} + 3 y = x^{2} + 2 e^{5 x}

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

x^{^{''}} + a^{2} x = a^{2} b - c

x^{^{''}} + x = sin (2 t), x (0) = 0, x^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} - 24 y = 5 e^{3 x} + x^{2}

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