What is the solution for y^{''}-y^'=3x^2 ?

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

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y^{''}-y^'=3x^2

y^{^{''}} - y^{^{'}} = 3 x^{2}

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + y = 4 x^{2} - 3 + \frac{e ^{x}}{x}

(d^{3} + 4 d) y = e^{x} + sin (x)

2 y^{^{''}} + 18 y^{^{'}} = 6 tan (3 x)

y^{^{''}} - 7 y^{^{'}} + 12 y = e^{3} x + e^{4} x

y^{^{''}} + 4 y^{^{'}} = 10 t + 1

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