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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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