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

The solution for y^{''}-3y^'=2x-2 is y=c_{1}e^{3x}+c_{2}-(x^2)/3+(4x)/9

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

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

y = c_{1} e^{3 x} + c_{2} - \frac{x ^{2}}{3} + \frac{4 x}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} - \frac{x ^{2}}{3} + \frac{4 x}{9}

y = c_{1} e^{3 x} + c_{2} - \frac{x ^{2}}{3} + \frac{4 x}{9}

y^{^{''}} + 4 y^{^{'}} + 5 y = - 15 x + e - x

y^{^{''}} + 2 y = t^{2} - 5, y (0) = 0, y^{^{'}} (0) = - 8

y^{^{''}} - y = - 15 t + 4

y^{^{''}} - y = - 15 t + 5

y^{^{''}} + 6 y^{^{'}} + 5 y = 12 e^{x}, y (0) = - 1, y^{^{'}} (0) = 7

What is the solution for y^{''}-3y^'=2x-2 ?
The solution for y^{''}-3y^'=2x-2 is y=c_{1}e^{3x}+c_{2}-(x^2)/3+(4x)/9