What is the solution for y^{''}+3y^'=9x ?

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

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y^{''}+3y^'=9x

y^{^{''}} + 3 y^{^{'}} = 9 x

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

Solution steps

y^{''} (x) + 3 y^{'} (x) = 9 x

Solve linear ODE:

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

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

\frac{d}{d x} (e^{(- 5 x^{2} - 2 x^{2})})

\frac{d}{d y} (\frac{y - 2 ( u y )}{y})

y^{^{''}} + 2 y = 18

\frac{1}{s ( s ^{2} + s + 5 )}

\frac{3}{x ^{2}}, at x = 1

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