Question

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 :

• A. $\frac{1}{2}$
• B. $\frac{3}{2}$
• C. $-\frac{3}{2}$
• D. $\frac{5}{2}$

Answer 1

Answer: (B)

$\frac{3}{2}$

Answer 2

The correct option is (B): $\frac{3}{2}$