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

The solution for y^{''}-3y^'=2x-3 is y=c_{1}e^{3x}+c_{2}-(x^2)/3+(7x)/9

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

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

y = c_{1} e^{3 x} + c_{2} - \frac{x ^{2}}{3} + \frac{7 x}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} - \frac{x ^{2}}{3} + \frac{7 x}{9}

y = c_{1} e^{3 x} + c_{2} - \frac{x ^{2}}{3} + \frac{7 x}{9}

y^{^{'''}} - 6 y^{^{''}} + 8 y^{^{'}} = sin (2 t) + t^{2} e^{t}

y^{^{''}} - 6 y^{^{'}} + 9 y = 15 x + 4

y^{^{''}} + y = 2 x \cdot e^{x} - 2 sin (x)

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

y^{^{''}} - 2 y^{^{'}} + y = 25 cos (2 x)

What is the solution for y^{''}-3y^'=2x-3 ?
The solution for y^{''}-3y^'=2x-3 is y=c_{1}e^{3x}+c_{2}-(x^2)/3+(7x)/9