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)

$1+\frac{1}{y}-\frac{e^{\frac{1}{y}}}{e}$

Answer 2

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