Consider the differential equation,

Question:

Consider the differential equation,

$y^{2} d x+\left(x-\frac{1}{y}\right) d y=0$. If value of $y$ is 1 when $x=1$, the the value of $x$ for which $y=2$, is :

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

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

  3. $\frac{5}{2}+\frac{1}{\sqrt{\mathrm{e}}}$

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


Correct Option: , 4

Solution:

$y^{2} d x+x d y=\frac{d y}{y}$

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

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

$e^{-\frac{1}{y}} \cdot x=\int e^{-\frac{1}{y}} \cdot \frac{1}{y^{3}} d y+C$

$x e^{-\frac{1}{y}}=e^{-\frac{1}{y}}+\frac{e^{-\frac{1}{y}}}{y}+C$

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

$x=\frac{3}{2}-\frac{1}{\sqrt{\mathrm{e}}}$ when $y=2$

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