What is the solution for y^{'''}-2y^{''}-y^'+2y=e^4x ?

The solution for y^{'''}-2y^{''}-y^'+2y=e^4x is y=c_{1}e^{2x}+c_{2}e^{-x}+c_{3}e^x+(e^4x)/2+(e^4)/4

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y^{'''}-2y^{''}-y^'+2y=e^4x

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = e^{4} x

y = c_{1} e^{2 x} + c_{2} e^{- x} + c_{3} e^{x} + \frac{e ^{4} x}{2} + \frac{e ^{4}}{4}

Solution steps

y^{'''} (x) - 2 y^{''} (x) - y^{'} (x) + 2 y = e^{4} x

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} + c_{3} e^{x} + \frac{e ^{4} x}{2} + \frac{e ^{4}}{4}

y = c_{1} e^{2 x} + c_{2} e^{- x} + c_{3} e^{x} + \frac{e ^{4} x}{2} + \frac{e ^{4}}{4}

What is the solution for y^{'''}-2y^{''}-y^'+2y=e^4x ?
The solution for y^{'''}-2y^{''}-y^'+2y=e^4x is y=c_{1}e^{2x}+c_{2}e^{-x}+c_{3}e^x+(e^4x)/2+(e^4)/4