Question

Consider the differential equation $y^{2}dx+(x-\frac{1}{y})dy=0.$ If y (1) =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 $1+\frac{1}{y}-\frac{e^{\frac{1}{y}}}{e}$

$y^{2}dx+(x-\frac{1}{y})dy=0$ & $y\left(1\right)=1$

$y^{2}dx+xdy=\frac{dy}{y}$

Multiply both sides by $e^{-\frac{1}{y}}$

$e^{-\frac{1}{y}}{y}^{2}dx+e^{-\frac{1}{y}}xdy=\frac{dy}{y}{e}^{-\frac{1}{y}}$

$e^{-\frac{1}{y}}dx+\left(\frac{e^{-\frac{1}{y}}}{y^2}\right)xdy=\frac{e^{-\frac{1}{y}}}{y^3}dy$

$e^{-\frac{1}{y}}=h\left(y\right)$ & $x=g\left(x\right)$

$g^{\prime }\left(x\right)=1$ & $h^{\prime }\left(y\right)=\frac{e^{-\frac{1}{y}}}{y^2}$

$h\left(y\right)g^{\prime }\left(x\right)dx+h^{\prime }\left(y\right)g\left(x\right)dy=\frac{e^{-\frac{1}{y}}}{y^3}dy$

Integrating both sides

$\int h\left(y\right)g^{\prime }\left(x\right)dx+h^{\prime }\left(y\right)g\left(x\right)dy=-\int e^{-\frac{1}{y}}(-\frac{1}{y})\frac{1}{y^2}dy$

Let $-\frac{1}{y}=v$

$\Rightarrow h\left(y\right)g\left(x\right)+c_1=-\int e^{u}udu$ $\frac{dy}{y^2}=du$

$\Rightarrow h\left(y\right)g\left(x\right)+c_1=-(u{e}^u-e^u)+c_2$ (using By parts)

$\Rightarrow e^{-\frac{1}{y}}x+c_1=e^{-\frac{1}{y}}+\frac{1}{y}{e}^{-\frac{1}{y}}+c_2$

$[c_2-c_1=c]$

$e^{-\frac{1}{y}}x=e^{-\frac{1}{y}}(1+\frac{1}{y})+c$

$x=1+\frac{1}{y}+c{e}^{\frac{1}{y}}$ as $f\left(1\right)=1$

$1+\frac{1}{1}+c{e}^{\frac{1}{1}}=1$

$c=-\frac{1}{e}$

Therefore $x=1+\frac{1}{y}-\frac{e^{\frac{1}{y}}}{e}$.