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

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

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

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

y = c_{1} e^{(- 2 + 3^{\frac{1}{2}}) x} + c_{2} e^{(- 2 - 3^{\frac{1}{2}}) x} + e^{- 2 x + 3^{\frac{1}{2}} x} \cdot \int \frac{e ^{\frac{3 x - 2 \cdot 3 ^{\frac{1}{2}} x}{2}}}{2 \cdot 3 ^{\frac{1}{2}} x ^{2}} d x + e^{- 2 x - 3^{\frac{1}{2}} x} \cdot \int - \frac{e ^{\frac{3 x + 2 \cdot 3 ^{\frac{1}{2}} x}{2}}}{2 \cdot 3 ^{\frac{1}{2}} x ^{2}} d x

Solution steps

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

Solve linear ODE:

y = c_{1} e^{(- 2 + 3^{\frac{1}{2}}) x} + c_{2} e^{(- 2 - 3^{\frac{1}{2}}) x} + e^{- 2 x + 3^{\frac{1}{2}} x} \cdot \int \frac{e ^{\frac{3 x - 2 \cdot 3 ^{\frac{1}{2}} x}{2}}}{2 \cdot 3 ^{\frac{1}{2}} x ^{2}} d x + e^{- 2 x - 3^{\frac{1}{2}} x} \cdot \int - \frac{e ^{\frac{3 x + 2 \cdot 3 ^{\frac{1}{2}} x}{2}}}{2 \cdot 3 ^{\frac{1}{2}} x ^{2}} d x

y = c_{1} e^{(- 2 + 3^{\frac{1}{2}}) x} + c_{2} e^{(- 2 - 3^{\frac{1}{2}}) x} + e^{- 2 x + 3^{\frac{1}{2}} x} \cdot \int \frac{e ^{\frac{3 x - 2 \cdot 3 ^{\frac{1}{2}} x}{2}}}{2 \cdot 3 ^{\frac{1}{2}} x ^{2}} d x + e^{- 2 x - 3^{\frac{1}{2}} x} \cdot \int - \frac{e ^{\frac{3 x + 2 \cdot 3 ^{\frac{1}{2}} x}{2}}}{2 \cdot 3 ^{\frac{1}{2}} x ^{2}} d x

2 y^{^{''}} - 2 y^{^{'}} - 4 y = 2 e^{3 t}

2 x^{^{''}} + 3 x^{^{'}} + 4 x = - 5 cos (t) + sin (t)

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{- x} \cdot cos (x)

y^{^{''}} + 6 y^{^{'}} + 5 y = 30

4 y^{^{''}} - 4 y^{^{'}} + y = 80 e^{\frac{t}{2}}

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