Thank you for reporting, we will resolve it shortly
Q.
If $2 x y d y=\left(x^2+y^2+1\right) d x, y(1)=0$ and $y\left(x_0\right)=\sqrt{3}$, then $x_0$ can be
Differential Equations
Solution:
$2 x y \frac{d y}{d x}=x^2+y^2+1$
Put $y^2=t$
$x \frac{d t}{d x}=x^2+t+1 $
$\frac{d t}{d x}-\frac{t}{x}=\frac{x^2+1}{x}$
$\text { I.F. }=\frac{1}{x}$
Hence $\frac{t}{x}=\int \frac{x^2+1}{x^2} d x=x-\frac{1}{x}+C$
$\therefore y ^2= x ^2-1, C =0$ as $y (1)=0$
Now, $y \left( x _0\right)=\sqrt{3} \Rightarrow 3= x _0^2-1 \Rightarrow x _0^2=4$
$\therefore x _0= \pm 2$