What is the solution for y^{''}-3y^'+3y^'-y=x-8e^x ?

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

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y^{''}-3y^'+3y^'-y=x-8e^x

y^{^{''}} - 3 y^{^{'}} + 3 y^{^{'}} - y = x - 8 e^{x}

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

Solution steps

y^{''} (x) - 3 y^{'} (x) + 3 y^{'} (x) - y = x - 8 e^{x}

Solve linear ODE:

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

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

y^{^{''}} + 100 y^{^{'}} + \frac{1}{0.0004} y = 30, y (0) = 0

y^{^{''''}} - 5 y^{^{'''}} + 4 y^{^{''}} = x

2 y^{^{''}} + y = 5 + cos (4 t)

y^{^{''}} + 4 y^{^{'}} + 3 y = x - 1

y^{^{'''}} - 6 y^{^{''}} = cos (4 x)

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