What is the solution for y^{''}-3y^'=e^{3x}+2x-1 ?

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

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

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

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

Solution steps

y^{''} (x) - 3 y^{'} (x) = e^{3 x} + 2 x - 1

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + y = 7 + 75 sin (2 x)

y^{^{''}} + 2 y^{^{'}} - 3 y = (- 4 x^{2} - 4 x + 2) e^{2 x}

y^{^{''''}} + 2 y^{^{''}} + y = 4 + cos (2 t)

y^{^{''}} + 9 y = 4, y (0) = 0, y^{^{'}} (0) = 0

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

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