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

The solution for y^{''}+3y^'=x+3 is y=c_{1}+c_{2}e^{-3x}+(x^2)/6+(8x)/9

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

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

y = c_{1} + c_{2} e^{- 3 x} + \frac{x ^{2}}{6} + \frac{8 x}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 3 x} + \frac{x ^{2}}{6} + \frac{8 x}{9}

y = c_{1} + c_{2} e^{- 3 x} + \frac{x ^{2}}{6} + \frac{8 x}{9}

y^{^{''}} + 3 y^{^{'}} = x^{2} + 9

y^{^{''}} + 9 y = 4 sec (3 x)

y^{^{''''}} + 20 y^{^{''}} + 100 = 0

x^{^{''}} - 4 x^{^{'}} + 4 x = 2 sin (2 t)

y^{^{''}} - 3 y^{^{'}} - 4 y = 1 - 8 x^{2} + 48 e^{- 4 x}

What is the solution for y^{''}+3y^'=x+3 ?
The solution for y^{''}+3y^'=x+3 is y=c_{1}+c_{2}e^{-3x}+(x^2)/6+(8x)/9