What is the solution for y^{''}-y^'-2x=4x^2 ?

The solution for y^{''}-y^'-2x=4x^2 is y=c_{1}e^x+c_{2}-(4x^3)/3-5x^2-10x

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

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

y = c_{1} e^{x} + c_{2} - \frac{4 x ^{3}}{3} - 5 x^{2} - 10 x

Solution steps

y^{''} (x) - y^{'} (x) - 2 x = 4 x^{2}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} - \frac{4 x ^{3}}{3} - 5 x^{2} - 10 x

y = c_{1} e^{x} + c_{2} - \frac{4 x ^{3}}{3} - 5 x^{2} - 10 x

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

3 y^{^{''}} - 9 y^{^{'}} = sin (2 x)

y^{^{''}} + 9 y = 3, y (0) = 0, y^{^{'}} (0) = 0

8 y^{^{''}} + 7 y^{^{'}} - y = 17

y^{^{''}} - 5 y^{^{'}} + 6 y = 5 e^{3 x} - 7 e^{5 x}

What is the solution for y^{''}-y^'-2x=4x^2 ?
The solution for y^{''}-y^'-2x=4x^2 is y=c_{1}e^x+c_{2}-(4x^3)/3-5x^2-10x