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

The solution for y^'=xy^2(1+x^2)^{-1/2},y(0)=2 is y=-2/(2(x^2+1)^{1/2)-3}

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

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

y = - \frac{2}{2 ( x ^{2} + 1 ) ^{\frac{1}{2}} - 3}

Solution steps

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

Solve separable ODE:

y = - \frac{2}{2 ( x ^{2} + 1 ) ^{\frac{1}{2}} - 3}

y = - \frac{2}{2 ( x ^{2} + 1 ) ^{\frac{1}{2}} - 3}

(4, 1), (6, 7)

\frac{\partial}{\partial x} ((1 + \frac{2}{y}) cos (x))

\frac{d}{d x} (3 cos (x^{2} - 1))

\int_{5}^{7} \frac{1}{4 + ( x - 5 ) ^{2}} d x

What is the solution for y^'=xy^2(1+x^2)^{-1/2},y(0)=2 ?
The solution for y^'=xy^2(1+x^2)^{-1/2},y(0)=2 is y=-2/(2(x^2+1)^{1/2)-3}