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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int x^{2} (3 - e^{x^{3}}) d x

x \to \infty lim (e^{- \frac{2}{x}})

\frac{2}{s}

\int_{1}^{2} 0 d x

\int_{- \infty}^{- 1} x^{- \frac{11}{5}} d x

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