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

The solution for y^{''}+2y^'+y=(1/6)(e^{-x}) is y=c_{1}e^{-x}+c_{2}xe^{-x}+(e^{-x}x^2)/(12)

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y^{''}+2y^'+y=(1/6)(e^{-x})

y^{^{''}} + 2 y^{^{'}} + y = (\frac{1}{6}) (e^{- x})

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{- x} x ^{2}}{12}

Solution steps

y^{''} (x) + 2 y^{'} (x) + y = (\frac{1}{6}) (e^{- x})

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{- x} x ^{2}}{12}

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{- x} x ^{2}}{12}

y^{^{''}} + 9 y = 7 x, y (0) = 1, y^{^{'}} (0) = 3

y^{^{''}} - 0.01 y = - 0.01 \cdot 20

y^{^{''}} + \frac{1}{2} y^{^{'}} + y = 4 sin (\frac{x}{2})

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

8 y^{^{''}} + y = t, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+2y^'+y=(1/6)(e^{-x}) ?
The solution for y^{''}+2y^'+y=(1/6)(e^{-x}) is y=c_{1}e^{-x}+c_{2}xe^{-x}+(e^{-x}x^2)/(12)