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

The solution for y^{''}-2y^'=x^3 is y=c_{1}e^{2x}+c_{2}-(x^4)/8-(x^3)/4-(3x^2)/8-(3x)/8

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

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

y = c_{1} e^{2 x} + c_{2} - \frac{x ^{4}}{8} - \frac{x ^{3}}{4} - \frac{3 x ^{2}}{8} - \frac{3 x}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} - \frac{x ^{4}}{8} - \frac{x ^{3}}{4} - \frac{3 x ^{2}}{8} - \frac{3 x}{8}

y = c_{1} e^{2 x} + c_{2} - \frac{x ^{4}}{8} - \frac{x ^{3}}{4} - \frac{3 x ^{2}}{8} - \frac{3 x}{8}

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 12 e^{x} - 12

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

\frac{d ^{2} y}{d t ^{2}} - 4 y = t cos (2 t)

y^{^{''}} + 25 y = 6 sin (x)

What is the solution for y^{''}-2y^'=x^3 ?
The solution for y^{''}-2y^'=x^3 is y=c_{1}e^{2x}+c_{2}-(x^4)/8-(x^3)/4-(3x^2)/8-(3x)/8