What is the solution for y^{''}-3y^'=x^3-6 ?

The solution for y^{''}-3y^'=x^3-6 is y=c_{1}e^{3x}+c_{2}-(x^4}{12}-(x^3)/9-(x^2)/9+\frac{52x)/(27)

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y^{''}-3y^'=x^3-6

y^{^{''}} - 3 y^{^{'}} = x^{3} - 6

y = c_{1} e^{3 x} + c_{2} - \frac{x ^{4}}{12} - \frac{x ^{3}}{9} - \frac{x ^{2}}{9} + \frac{52 x}{27}

Solution steps

y^{''} (x) - 3 y^{'} (x) = x^{3} - 6

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} - \frac{x ^{4}}{12} - \frac{x ^{3}}{9} - \frac{x ^{2}}{9} + \frac{52 x}{27}

y = c_{1} e^{3 x} + c_{2} - \frac{x ^{4}}{12} - \frac{x ^{3}}{9} - \frac{x ^{2}}{9} + \frac{52 x}{27}

\frac{d}{d x} (15 cos (3 x))

\frac{d}{d x} (x^{4} - 1)

\frac{d}{d x} (\frac{3 x}{x - 2})

\int \frac{1}{16 e ^{- 4 x} + e ^{4 x}} d x

sin ((π x)^{4})

What is the solution for y^{''}-3y^'=x^3-6 ?
The solution for y^{''}-3y^'=x^3-6 is y=c_{1}e^{3x}+c_{2}-(x^4}{12}-(x^3)/9-(x^2)/9+\frac{52x)/(27)