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 $x=1$, then the value of $x$ for which $y=2$, is :

  1. (1) $\frac{5}{2}+\frac{1}{\sqrt{e}}$

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

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

  4. (4) $\frac{3}{2}-\sqrt{e}$.


Correct Option: , 2

Solution:

Consider the differential equation,

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

$\Rightarrow \frac{d x}{d y}+\left(\frac{1}{y^{2}}\right) x=\frac{1}{y^{3}}$

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

$\therefore \quad x . e^{-\frac{1}{y}}=\int e^{-\frac{1}{y}} \frac{1}{y^{3}} d y+c$

Put $-\frac{1}{y}=u \Rightarrow \frac{1}{y^{2}} d y=d u$

$\Rightarrow x . e^{-\frac{1}{y}}=-\int u e^{u} d u+c=-u e^{u}+e^{u}+c$

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

At $y=1, x=1$

$1=2+c e \Rightarrow c=-\frac{1}{e} \Rightarrow x=\left(1+\frac{1}{y}\right)-\frac{1}{e} e^{\frac{1}{y}}$

On putting $y=2$, we get $x=\frac{3}{2}-\frac{1}{\sqrt{e}}$

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