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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + y = cos^{2} (x)

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

y^{^{''}} - 2 y + 5 y = 1, + e^{- t} y (0) = 0, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} + y = x cos (2 x)

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

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