Let $y=y(x)$ be a solution curve of the differential equation, $(1-x^{2} y^{2}) dx=y dx+x dy$. If the line $\mathrm{x}=1$ intersects the curve $\mathrm{y}=\mathrm{y}(\mathrm{x})$ at $\mathrm{y}=2$ and the line $\mathrm{x}=2$ intersects the curve $\mathrm{y}=\mathrm{y}(\mathrm{x})$ at $\mathrm{y}=\alpha$, then a value of $\alpha$ is

(D) $\frac{1+3 e^{2}}{2(3 e^{2}-1)}$

(A) $\frac{3 \mathrm{e}^{2}}{2(3 \mathrm{e}^{2}-1)}$

(B) $\frac{3 \mathrm{e}^{2}}{2(3 \mathrm{e}^{2}+1)}$

(C) $\frac{1-3 e^{2}}{2(3 e^{2}+1)}$

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Differential Equations

Let

y = y (x)

be a solution curve of the differential equation,

(1 - x^{2} y^{2}) d x = y d x + x d y

.

If the line

x = 1

intersects the curve

y = y (x)

y = 2

and the line

x = 2

intersects the curve

y = y (x)

y = α

, then a value of

α

Answer: (D)

The right side satisfies

y d x + x d y = d (x y)

, so the equation becomes

(1 - (x y)^{2}) d x = d (x y)

, i.e.

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

. Integrating with

u = x y

x = \frac{1}{2} ln \frac{1 + x y}{1 - x y} + C

. Using

y = 2

x = 1

gives

1 = \frac{1}{2} ln 3 + C

. At

x = 2

with

y = α

2 = \frac{1}{2} ln \frac{1 + 2 α}{1 - 2 α} + 1 - \frac{1}{2} ln 3

, hence

\frac{1 + 2 α}{1 - 2 α} = 3 e^{2}

. Taking the branch

\frac{1 + 2 α}{1 - 2 α} = - 3 e^{2}

gives

α = \frac{1 + 3 e ^{2}}{2 ( 3 e ^{2} - 1 )}

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Single correct option