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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - y^{^{'}} - 6 y = e^{3 x} + 30

y^{^{''}} + 4 y = 1

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

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{1 + e ^{2 x}}

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