Question

Consider the differential equation, $y^{2}dx+(x-\frac{1}{y})dy=0.$ If value of $y$ is $1$ when $x=1,$ then the value of $x$ for which $y=2,$ is:

• A. $\frac{3}{2}-\sqrt{e}$
• B. $\frac{5}{2}+\frac{1}{\sqrt{e}}$
• C. $\frac{3}{2}-\frac{1}{\sqrt{e}}$
• D. $\frac{1}{2}+\frac{1}{\sqrt{e}}$

Answer 1

The correct option is C $\frac{3}{2}-\frac{1}{\sqrt{e}}$

$y^{2}dx+(x-\frac{1}{y})dy=0$
$\Rightarrow \frac{dx}{dy}+\frac{x}{y^2}=\frac{1}{y^3}$
Integrating Factor
I.F. $=e^{\int \frac{1}{y^2}dy}=e^{-\frac{1}{y}}$
Therefore differential equation becomes
$\Rightarrow x{e}^{-\frac{1}{y}}=\int e^{-\frac{1}{y}}\cdot \frac{1}{y^3}dy$
Let $-\frac{1}{y}=t\Rightarrow dt=\frac{1}{y^2}dy$
$\Rightarrow x\cdot e^t=\int -t\cdot e^{t}dt$
$\Rightarrow x\cdot e^t=-t\cdot e^t+e^t+C$
$\Rightarrow x\cdot e^{-\frac{1}{y}}=(\frac{1}{y}+1)\cdot e^{-\frac{1}{y}}+C$
As $y\left(1\right)=1$
$\therefore C=-\frac{1}{e}$
$\Rightarrow x\cdot e^{-\frac{1}{y}}=(\frac{1}{y}+1)\cdot e^{-\frac{1}{y}}-\frac{1}{e}$
$\Rightarrow x=\frac{1}{y}+1-\frac{1}{e}\cdot e^{\frac{1}{y}}$

At $y=2$
$\Rightarrow x=\frac{3}{2}-\frac{1}{\sqrt{e}}$