# If y=y(x) is the solution curve

Question:

If $y=y(x)$ is the solution curve of the differential equation $x^{2} d y+\left(y-\frac{1}{x}\right) d x=0 \quad ; x>0$ and $y(1)=1$, then $y\left(\frac{1}{2}\right)$ is equal to :

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

2. $3+\frac{1}{\sqrt{\mathrm{e}}}$

3. $3+\mathrm{e}$

4. $3-e$

Correct Option: , 4

Solution:

$x^{2} d y+\left(y-\frac{1}{x}\right) d x=0: x>0, y(1)=1$

$x^{2} d y+\frac{(x y-1)}{x} d x=0$

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

$\frac{d y}{d x}=\frac{1-x y}{x^{3}}$

$\frac{d y}{d x}=\frac{1}{x^{3}}-\frac{y}{x^{2}}$

$\frac{d y}{d x}=\frac{1}{x^{2}} \cdot y=\frac{1}{x^{3}}$

If $e^{\int \frac{1}{x^{2}} d x}=e^{-\frac{1}{x}}$

$y e^{-\frac{1}{x}}=\int \frac{1}{x^{3}} e^{-\frac{1}{x}}$

$y e^{-\frac{1}{x}}=e^{-x}\left(1+\frac{1}{x}\right)+C$

$1 . e^{-1}=e^{-1}(2)+C$

$C=-e^{-1}=-\frac{1}{e}$

$y e^{-\frac{1}{x}}=e^{-\frac{1}{x}}\left(1+\frac{1}{x}\right)-\frac{1}{e}$

$y\left(\frac{1}{2}\right)=3-\frac{1}{e} \times e^{2}$

$y\left(\frac{1}{2}\right)=3-e$