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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} - 2 y = - 4 x^{2} - 6 x + 1

y^{^{''}} - 4 y^{^{'}} + 5 y = 10 x - 3

y^{^{''}} - 6 y^{^{'}} + 8 y = x e^{x}

- y^{^{''}} - 4 y = t^{2}

y^{^{''}} + 6 y^{^{'}} + 73 y = 8 cos (4 x)

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