Question

$y^2\frac{dx}{dy}+x-\frac{1}{y}=0$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ {y}^2\frac{dx}{dy}+x-\frac{1}{y}=0 \\ \Rightarrow y^2\frac{dx}{dy}+x=\frac{1}{y} \\ \Rightarrow \frac{dx}{dy}+\frac{1}{y^2}x=\frac{1}{y^3}.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dx}{dy}+Px=Q \\ \mathrm{where} \\ P=\frac{1}{y^2} \\ Q=\frac{1}{y^3} \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdy} \\ =e^{\int \frac{1}{y^2}dy} \\ =e^{\frac{-1}{y}} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}{e}^{\frac{-1}{y}},\mathrm{we}\mathrm{get} \\ {e}^{\frac{-1}{y}}\left(\frac{dx}{dy}+x\frac{1}{y^2}\right)=e^{\frac{-1}{y}}\frac{1}{y^3} \\ \Rightarrow e^{\frac{-1}{y}}\frac{dx}{dy}+x\frac{1}{y^2}{e}^{\frac{-1}{y}}=e^{\frac{-1}{y}}\frac{1}{y^3} \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}y,\mathrm{we}\mathrm{get} \\ x{e}^{\frac{-1}{y}}=\int e^{\frac{-1}{y}}\frac{1}{y^3}dy+C \\ \Rightarrow x{e}^{\frac{-1}{y}}=I+C.....\left(2\right) \\ \mathrm{where} \\ I=\int e^{\frac{-1}{y}}\frac{1}{y^3}dy \\ \mathrm{Putting}t=\frac{1}{y},\mathrm{we}\mathrm{get} \\ dt=-\frac{1}{y^2}dy \\ \therefore I=-\int \underset{\mathrm{I}}{t}\underset{\mathrm{II}}{e^{-t}}dt \\ =-t\int e^{-t}dt+\int \left[\frac{d}{dt}\left(t\right)\int e^{-t}dt\right]dt \\ =t{e}^{-t}+e^{-t} \\ =\left(t+1\right)e^{-t} \\ =\left(\frac{1}{y}+1\right)e^{-\frac{1}{y}} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}I\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ x{e}^{\frac{-1}{y}}=\left(\frac{1}{y}+1\right)e^{-\frac{1}{y}}+C \\ \Rightarrow x=\left(\frac{y+1}{y}\right)+C{e}^{\frac{1}{y}} \\ \mathrm{Hence},x=\left(\frac{y+1}{y}\right)+C{e}^{\frac{1}{y}}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$