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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 16 y = 56 cos (4 t)

x^{^{''}} + 5 x^{^{'}} + 17 x = 12 cos (3 t)

y^{^{''}} + 2 y^{^{'}} + y = e^{- x} (2 x + 5)

y^{^{''}} + y = 3 sin (2 t), y (0) = 12, y^{^{'}} (0) = 1

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

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