Question

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

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

Answer 1

Answer: (C)

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