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}

\frac{d ^{2} y}{d x ^{2}} + y = (sec (x))^{3}

y^{^{''}} - 2 y^{^{'}} - 3 y = 2 e^{5 t}

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

(D^{2} - 4 D + 4) y = 8 (x^{2} + e^{2 x} + sin (2 x))

y^{^{''}} + y = cos (x - \frac{π}{2}), y (0) = 0, y^{^{'}} (0) = 0

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