Consider the differential equation y2dx+x−1ydy=0 if value of y is 1 when x = 1, then the value of x for which y = 2, is

# Consider the differential equation ${y}^{2}dx+\left(x-\frac{1}{y}\right)dy=0$ if value of $y$ is 1 when  then the value offor which , is

1. A

$\frac{1}{2}+\frac{1}{\sqrt{e}}$

2. B

$\frac{5}{2}+\frac{1}{\sqrt{e}}$

3. C

$\frac{3}{2}-\sqrt{e}$

4. D

$\frac{3}{2}-\frac{1}{\sqrt{e}}$

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### Solution:

Given differential equation is

${y}^{2}dx+\left(x-\frac{1}{y}\right)dy=0$

$⇒\frac{dx}{dy}+\frac{x}{{y}^{2}}=\frac{1}{{y}^{3}}$ which is a linear differential equation.

I.F. $={e}^{\int \frac{1}{{y}^{2}}dy}={e}^{-\frac{1}{y}}$

Required solution is $x\cdot {e}^{-\frac{1}{y}}=\int {e}^{-\frac{1}{y}}\frac{1}{{y}^{3}}dy+C$

Let $I=\int {e}^{-\frac{1}{y}}\frac{1}{{y}^{3}}dy$ Putting $-\frac{1}{y}=t⇒\frac{dy}{{y}^{2}}=dt$

Equation of curve is $x=1+\frac{1}{y}-{e}^{\frac{1}{y}-1}$

At $y=2,x=1+\frac{1}{2}-{e}^{\frac{1}{2}-1}=\frac{3}{2}-\frac{1}{\sqrt{e}}$

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