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

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

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

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

y = e^{c_{1}} x^{2} + e^{c_{1}}

Solution steps

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

Solve separable ODE:

y = e^{c_{1}} x^{2} + e^{c_{1}}

y = e^{c_{1}} x^{2} + e^{c_{1}}

\frac{d y}{d x} = - 4 e^{x + y}

y^{2} - x^{2} y^{^{'}} = 0

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

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

x^{4} \frac{d y}{d x} + y^{5} = 0

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