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

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

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

y^{^{'}} + \frac{1 - x}{2 x} y = 0

y = e^{- \frac{1}{2} (- x + l n (x)) + c_{1}}

Solution steps

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

Solve separable ODE:

y = e^{- \frac{1}{2} (- x + l n (x)) + c_{1}}

y = e^{- \frac{1}{2} (- x + l n (x)) + c_{1}}

f (x) = 2 x + \frac{1936}{x}

f (x) = 110 x - 16 x^{2}, at x = 4

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

\frac{1}{x + 8}

\frac{d}{d x} (\frac{e ^{x}}{x ^{2} + 7})

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