Question

Consider the differential equation
$y^{2}dx+(x-\frac{1}{y})dy=0$.
$\text{If}y\left(1\right)=1$, then $x$ is given by :

• A. $4-\frac{2}{y}-\frac{e^{\frac{1}{y}}}{e}$
  
• B. $3-\frac{1}{y}-\frac{e^{\frac{1}{y}}}{e}$
• C. $1+\frac{1}{y}-\frac{e^{\frac{1}{y}}}{e}$
• D. $1-\frac{1}{y}-\frac{e^{\frac{1}{y}}}{e}$

Answer 1

The correct option is C $1+\frac{1}{y}-\frac{e^{\frac{1}{y}}}{e}$

$\frac{dx}{dy}+\frac{x}{y^2}=\frac{1}{y^3}$
I.F.$=e^{\int \frac{1}{y^2}dy}=e^{-\frac{1}{y}}$
So $x.e^{-\frac{1}{y}}=\int \frac{1}{y^3}{e}^{-\frac{1}{y}}dy$
Let $\frac{-1}{y}=t\Rightarrow \frac{1}{y^2}dy=dt$
$\Rightarrow I=-\int t{e}^{t}dt=e^t-t{e}^t=e^{-\frac{1}{y}}+\frac{1}{y}{e}^{-\frac{1}{y}}+c$
$\Rightarrow x=1+\frac{1}{y}+c.e^{\frac{1}{y}}$
Since y(1)=1
$\therefore c=-\frac{1}{e}$
$\Rightarrow x=1+\frac{1}{y}-\frac{1}{e}.e^{\frac{1}{y}}$