Question

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

Answer 1

We have,
$y^2+\left(x+\frac{1}{y}\right)\frac{dy}{dx}=0$
$$\begin{gathered} \Rightarrow \frac{dy}{dx}=-\frac{y^3}{xy+1} \\ \Rightarrow \frac{dx}{dy}=-\frac{xy+1}{y^3} \\ \Rightarrow \frac{dx}{dy}=-\frac{x}{y^2}-\frac{1}{y^3} \\ \Rightarrow \frac{dx}{dy}+\frac{x}{y^2}=-\frac{1}{y^3} \end{gathered}$$
$$\begin{gathered} \mathrm{Comparing}\mathrm{with}\frac{dx}{dy}+Px=Q,\mathrm{we}\mathrm{get} \\ P=\frac{1}{y^2} \\ Q=-\frac{1}{y^3} \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{\int \frac{1}{y^2}dy}=e^{-\frac{1}{y}} \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ x\times {\mathrm{e}}^{-\frac{1}{\mathrm{y}}}=\int -e^{-\frac{1}{y}}\frac{1}{y^3}dy+C \\ \Rightarrow x{\mathrm{e}}^{-\frac{1}{\mathrm{y}}}=I+C.....\left(1\right) \\ \mathrm{Now}, \\ I=\int -e^{-\frac{1}{y}}\frac{1}{y^3}dy \\ \mathrm{Putting}t=\frac{1}{y},\mathrm{we}\mathrm{get} \\ dt=\frac{-1}{y^2}dy \\ \therefore I=\int \underset{\mathrm{I}}{t}\times \underset{\mathrm{II}}{{\mathrm{e}}^{-t}}dt \\ =t\times \int e^{-t}dt-\int \left(\frac{dt}{dt}\times \int e^{-t}dt\right)dt \\ =-t{e}^t+\int e^{-t}dt \\ =-t{e}^{-t}-e^{-t} \\ \therefore I=-\frac{1}{y}{e}^{-\frac{1}{y}}-e^{-\frac{1}{y}}=-e^{-\frac{1}{y}}\left(1+\frac{1}{y}\right) \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}I\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ x{\mathrm{e}}^{-\frac{1}{\mathrm{y}}}=-e^{-\frac{1}{y}}\left(1+\frac{1}{y}\right)+C \\ \Rightarrow x=-\left(1+\frac{1}{y}\right)+C{e}^{\frac{1}{y}} \end{gathered}$$