Thank you for reporting, we will resolve it shortly
Q.
Let the solution curve of the differential equation $x d y=\left(\sqrt{x^2+y^2}+y\right) d x, x>0$, intersect the line $x =1$ at $y =0$ and the line $x=2$ at $y=\alpha$. Then the value of $\alpha$ is :
$ x d y=\left(\sqrt{x^2+y^2}+y\right) d x $
$ x d y-y d x=\sqrt{x^2+y^2} d x $
$ \frac{x d y-y d x}{x^2}=\sqrt{1+\frac{y^2}{x^2}} \cdot \frac{d x}{x} $
$ \frac{d(y / x)}{\sqrt{1+\left(\frac{y}{x}\right)^2}}=\frac{d x}{x} $
$ \ln \left(\frac{y}{x}+\sqrt{\left(\frac{y}{x}\right)^2+1}\right)=\ln x+R $
$ \frac{y+\sqrt{y^2+x^2}}{x}=c x$
$ y +\sqrt{ y ^2+ x ^2}= cx ^2$
$ x =1, y =0 \Rightarrow 0+1= C \Rightarrow C =1 $
Curve is $y +\sqrt{ x ^2+ y ^2}= x ^2$
$x =2, y =\alpha$
$ 2+\sqrt{4+\alpha^2}=4 $
$ 4+\alpha^2=16+\alpha^2=8 \alpha $
$\alpha=\frac{3}{2}$