What is the solution for y^{''}-y^'=x^3 ?

The solution for y^{''}-y^'=x^3 is y=c_{1}e^x+c_{2}-(x^4)/4-x^3-3x^2-6x

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y^{''}-y^'=x^3

y^{^{''}} - y^{^{'}} = x^{3}

y = c_{1} e^{x} + c_{2} - \frac{x ^{4}}{4} - x^{3} - 3 x^{2} - 6 x

Solution steps

y^{''} (x) - y^{'} (x) = x^{3}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} - \frac{x ^{4}}{4} - x^{3} - 3 x^{2} - 6 x

y = c_{1} e^{x} + c_{2} - \frac{x ^{4}}{4} - x^{3} - 3 x^{2} - 6 x

What is the solution for y^{''}-y^'=x^3 ?
The solution for y^{''}-y^'=x^3 is y=c_{1}e^x+c_{2}-(x^4)/4-x^3-3x^2-6x