What is the solution for x(x-1)y+xy^'+2y=0 ?

The solution for x(x-1)y+xy^'+2y=0 is y=(e^{(-x^2+2x+c_{1})/2})/(x^2)

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x(x-1)y+xy^'+2y=0

x (x - 1) y + x y^{^{'}} + 2 y = 0

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

Solution steps

x (x - 1) y + x y^{'} (x) + 2 y = 0

Solve separable ODE:

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

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

\frac{d}{d x} (\frac{1}{x} + \frac{1}{y}) = 1

\frac{d}{d x} (e^{x} sin (y)) = 0

\frac{d y}{d t} = \frac{t}{t ^{2} y - y}

y^{^{'}} = (cos (x)) sin (y)

\frac{d}{d θ} (y (θ)) y (θ) = cos (e^{- 5 θ^{5}})

What is the solution for x(x-1)y+xy^'+2y=0 ?
The solution for x(x-1)y+xy^'+2y=0 is y=(e^{(-x^2+2x+c_{1})/2})/(x^2)