Mathematics - Differential Equation Question with Solution | TestHub

$\begin{array}{rl}& y^{2}dx+\left(x-\frac{1}{y}\right)dy=0\\ & y^{2}dx=\left(\frac{1}{y}-x\right)dy\\ & \Rightarrow y^2\frac{dx}{dy}=\frac{1}{y}-x\\ & \Rightarrow \frac{dx}{dy}+\frac{x}{y^2}=\frac{1}{y^3}\\ & \text{ I.F. }=e^{\int \frac{1}{y^2}dy}=e^{\frac{-1}{y}}\end{array}$
$\therefore$ Solution is
$x{e}^{\frac{-1}{y}}=\int e^{-\frac{1}{y}}\times \frac{1}{y^3}dy+C$
Let $\frac{-1}{y}=t$
$\begin{array}{rl}& \Rightarrow \frac{1}{y^2}dy=dt\\ & \Rightarrow x{e}^{-\frac{1}{y}}=-\int e^{t}tdt+C\\ & \Rightarrow x{e}^{-\frac{1}{y}}=-e^t(t-1)+C\\ & \Rightarrow x{e}^{-\frac{1}{y}}=-e^{\frac{-1}{y}}\left(\frac{-1}{y}-1\right)+C\end{array}$
$x(1)=1$
$\Rightarrow e^{-1}=-e^{-1}(-2)+C$
$\Rightarrow C=-e^{-1}$
$\Rightarrow x=\frac{1}{y}+1-e^{-1+\frac{1}{y}}$
$x\left(\frac{1}{2}\right)=3-e$