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

The solution for y^{''}+6y^'=1-3x+x^2 is y=c_{1}+c_{2}e^{-6x}+(x^3)/(18)-(5x^2)/(18)+(7x)/(27)

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

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

y = c_{1} + c_{2} e^{- 6 x} + \frac{x ^{3}}{18} - \frac{5 x ^{2}}{18} + \frac{7 x}{27}

Solution steps

y^{''} (x) + 6 y^{'} (x) = 1 - 3 x + x^{2}

Solve linear ODE:

y = c_{1} + c_{2} e^{- 6 x} + \frac{x ^{3}}{18} - \frac{5 x ^{2}}{18} + \frac{7 x}{27}

y = c_{1} + c_{2} e^{- 6 x} + \frac{x ^{3}}{18} - \frac{5 x ^{2}}{18} + \frac{7 x}{27}

y^{^{''}} + 4 y = 3 cos (2 x)

x^{^{''}} + 8 x^{^{'}} + 16 x = 8 sin (4 t)

y^{^{'''}} + y^{^{''}} = 5 x^{2}

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{3}

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 9 y = (x + 1) e^{x}

What is the solution for y^{''}+6y^'=1-3x+x^2 ?
The solution for y^{''}+6y^'=1-3x+x^2 is y=c_{1}+c_{2}e^{-6x}+(x^3)/(18)-(5x^2)/(18)+(7x)/(27)