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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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