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Q.
The solution of $\frac{d y}{d x}=\frac{x^{2}+y^{2}+1}{2 x y}$ satisfying $y(1)=1$ is given by
Differential Equations
Solution:
Rewriting the given equation as
$2 x y \frac{d y}{d x}-y^{2}=1+x^{2}$
$\Rightarrow 2 y \frac{d y}{d x}-\frac{1}{x} y^{2}=\frac{1}{x}+x$.
Putting $y^{2}=u,$ we have
$\frac{d u}{d x}-\frac{1}{x} u=\frac{1}{x}+x .$
The $I.F$. of this equation is $e^{-\int \frac{1}{x} d x}=\frac{1}{x}$
$u. \frac{1}{x}=\int\left(\frac{1}{x^{2}}+1\right) d x=-\frac{1}{x}+x+C$
$\Rightarrow y^{2}=\left(x^{2}-1\right)+C x$.
Since $y(1)=1$ so $C=1, y^{2}=x(1+x)-1$
which represents a system of hyperbola.