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 $\mathrm{x}=1$ at $\mathrm{y}=0$ and the line $\mathrm{x}=2$ at $\mathrm{y}=\alpha$. Then the value of $\alpha$ is :

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Differential Equations

Let the solution curve of the differential equation

x d y = (x^{2} + y^{2} + y) d x, x > 0

, intersect the line

x = 1

y = 0

and the line

x = 2

y = α

. Then the value of

α

is :

Answer: (B)

Rewrite the differential equation as

x d y - y d x = x^{2} + y^{2} d x

. Dividing by

x^{2}

\frac{d ( y / x )}{1 + ( y / x ) ^{2}} = \frac{d x}{x}

. Integrating:

ln (\frac{y}{x} + (y / x)^{2} + 1) = ln x + R

, which gives

y + y^{2} + x^{2} = c x^{2}

. Using the initial condition

x = 1, y = 0

0 + 1 = c

, so

c = 1

. The curve is

y + x^{2} + y^{2} = x^{2}

. At

x = 2, y = α

α + 4 + α^{2} = 4

. Solving:

4 + α^{2} = 4 - α

. Squaring:

4 + α^{2} = 16 - 8 α + α^{2}

, giving

8 α = 12

, so

α = \frac{3}{2}

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option