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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(D - 1) y = 2 sin (5 t), y (0) = 0

x^{^{''}} - 3 x^{^{'}} + x = - e^{t}

y^{^{''}} - 4 y^{^{'}} = 1

y^{^{''}} + 5 y = e^{2 x} - x - x^{2} + x^{3}

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 6 e^{x}

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