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

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

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

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

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

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}

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{x} - 12

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

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

y^{^{''}} - 6 y^{^{'}} + 5 y = - 9 e^{2 x}

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