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