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

The solution for y^{''}+3y^'=x is y=c_{1}+c_{2}e^{-3x}+(x^2)/6-x/9

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y^{''}+3y^'=x

y^{^{''}} + 3 y^{^{'}} = x

y = c_{1} + c_{2} e^{- 3 x} + \frac{x ^{2}}{6} - \frac{x}{9}

Solution steps

y^{''} (x) + 3 y^{'} (x) = x

Solve linear ODE:

y = c_{1} + c_{2} e^{- 3 x} + \frac{x ^{2}}{6} - \frac{x}{9}

y = c_{1} + c_{2} e^{- 3 x} + \frac{x ^{2}}{6} - \frac{x}{9}

y^{^{'''}} + 3 y^{^{''}} + 3 y^{^{'}} + y = x e^{- x}

y^{^{''}} - 12 y^{^{'}} + 36 y = 18 e^{6 x}

y^{^{''}} - 5 y^{^{'}} - 6 y = e^{x}

x^{^{''}} + 3 x^{^{'}} + 2 x = 1, x (0) = 0

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

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