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 ) at y = 2 and the line x = 2 intersects the curve y = y ( x ) at y = , then a value of

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Solution

Q.
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 $y = y (x)$ at $y = 2$ and the line $x = 2$ intersects the curve $y = y (x)$ at $y = α$ , then a value of $α$ is [2023]

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

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

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

4 $\frac{1 + 3 e^{2}}{2 (3 e^{2} - 1)}$

Ans.
(4)

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

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

$\int d x = \int \frac{d (x y)}{1 - {(x y)}^{2}}$ $x = \frac{1}{2} \ln$ $| \frac{1 + x y}{1 - x y} |$ $+ C$

$We are given y (1) = 2 and y (2) = α$

$Put x = 1, y = 2 : 1 = \frac{1}{2} \ln$ $| \frac{1 + 2}{1 - 2} |$ $+ C$

$C = 1 - \frac{1}{2} \ln 3$

$Now put x = 2, y = α$

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

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

$\Rightarrow$ $| \frac{1 + 2 α}{1 - 2 α} |$ $= 3 e^{2}$ $\Rightarrow \frac{1 + 2 α}{1 - 2 α} = 3 e^{2}, \frac{1 + 2 α}{1 - 2 α} = - 3 e^{2}$

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

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

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