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

The solution for y^{''}+3y^'=2x-3 is y=c_{1}+c_{2}e^{-3x}+(x^2)/3-(11x)/9

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

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

y = c_{1} + c_{2} e^{- 3 x} + \frac{x ^{2}}{3} - \frac{11 x}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 3 x} + \frac{x ^{2}}{3} - \frac{11 x}{9}

y = c_{1} + c_{2} e^{- 3 x} + \frac{x ^{2}}{3} - \frac{11 x}{9}

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

\frac{d ^{2} y}{d x ^{2}} + 5 \frac{d y}{d x} + 6 = 0

y^{^{'''}} - y = 1

y^{^{''}} + 7 y^{^{'}} + 12 y = 4 e^{5 t}, y (0) = 2, y^{^{'}} (0) = 0

100 y^{^{''}} - 60 y^{^{'}} + 13 y + 30 = 0

What is the solution for y^{''}+3y^'=2x-3 ?
The solution for y^{''}+3y^'=2x-3 is y=c_{1}+c_{2}e^{-3x}+(x^2)/3-(11x)/9