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

The solution for y^{''''}-2y^{''}+y=e^x is y=c_{1}e^x+c_{2}xe^x+c_{3}e^{-x}+c_{4}xe^{-x}+(e^xx^2)/8

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} e^{- x} + c_{4} x e^{- x} + \frac{e ^{x} x ^{2}}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} e^{- x} + c_{4} x e^{- x} + \frac{e ^{x} x ^{2}}{8}

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} e^{- x} + c_{4} x e^{- x} + \frac{e ^{x} x ^{2}}{8}

What is the solution for y^{''''}-2y^{''}+y=e^x ?
The solution for y^{''''}-2y^{''}+y=e^x is y=c_{1}e^x+c_{2}xe^x+c_{3}e^{-x}+c_{4}xe^{-x}+(e^xx^2)/8