Answer
\[y(x)=\sqrt{\frac{C+x^2}{x}}\]
Work Step by Step
$(y^2-2x)dx+2xydy=0$ ___(1)
Here,
$M(x,y)=y^2-2x$
$N(x,y)=2xy$
$M_y=2y\;\;\;\;,\;\;\;\;N_x=2y$
$M_y=N_x$
$\Rightarrow$ (1) is exact differential equation
Therefore there exists a potential function $\phi$ such that
$\frac{\partial\phi}{\partial x}=y^2-2x$ ___(2)
$\frac{\partial\phi}{\partial y}=2xy$ ___(3)
Integrating (2) with respect to $x$ holding $y$ fixed
$\phi =y^2x-x^2+h(y)$ ___(4)
Where $h(y)$ is arbitrary function of $y$
Differentiate (4) with respect to $y$ holding $x$ fixed
$\frac{\partial\phi}{\partial y}=2xy+\frac{dh}{dy}$ ___(5)
From (4) and (5)
$\frac{dh}{dy}=0 \Rightarrow h(y)=C_1$
$C_1$ is constant
From (4)
$\phi(x,y)=y^2x-x^2+C_1$
One parameter family is solutions is given by $\phi(x,y)=C_2$
$\Rightarrow y(x)=\sqrt{\frac{C+x^2}{x}}$
Where, $C=C_2-C_1$.
Hence general solution of (1) is $y(x)=\sqrt{\frac{C+x^2}{x}}$.
