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

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

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

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

y = \int e^{- 2 (- \frac{1}{x} - a r c t a n (x)) + c_{1}} d x

Solution steps

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

Solve using substitution:

y = \int e^{- 2 (- \frac{1}{x} - a r c t a n (x)) + c_{1}} d x

y = \int e^{- 2 (- \frac{1}{x} - a r c t a n (x)) + c_{1}} d x

\frac{d}{d x} (2 cos (x) - 4 sin (x))

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\int \frac{x}{7 - x - x ^{2}} d x

\int \frac{1}{169 - 16 x ^{2}} d x

\frac{d}{d x} (e^{x a})

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