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

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

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y^'+(1/x-1)y=-2/x

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (5 (2 - x^{3})^{4})

\frac{\partial}{\partial x} (- 3 y^{5} sin (3 x))

\frac{7 s + 2}{s ^{2} + 6 s + 34}

n \to \infty lim (e^{\frac{1}{n} l n (n)})

\frac{d}{d x} (2 x^{3} - 3)

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