Question

Find the particular solution of the differential equation
(1 – y²) (1 + log x) dx + 2xy dy = 0, given that y = 0 when x = 1.

Answer 1

Given:
$$\begin{gathered} \left(1-y^2\right)\left(1+\log x\right)\mathrm{d}x+2xy\mathrm{d}y=0 \\ \Rightarrow \left(1-y^2\right)\left(1+\log x\right)\mathrm{d}x=-2xy\mathrm{d}y \\ \Rightarrow \left(\frac{1+\log x}{2x}\right)\mathrm{d}x=-\left(\frac{y}{1-y^2}\right)\mathrm{d}y.....\left(1\right) \\ \mathrm{Let}: \\ 1+\log x=t \\ \mathrm{and} \\ \left(1-y^2\right)=p \\ \Rightarrow \frac{1}{x}\mathrm{d}x=\mathrm{d}t\mathrm{and}-2y\mathrm{d}y=\mathrm{d}p \\ \mathrm{Therefore},\left(1\right)\mathrm{becomes} \end{gathered}$$

$$\begin{gathered} \int \frac{t}{2}\mathrm{d}t=\int \frac{1}{2p}\mathrm{d}p \\ \Rightarrow \frac{t^2}{4}=\frac{\log p}{2}+C.....\left(2\right) \\ \mathrm{Substituting}\mathrm{the}\mathrm{values}\mathrm{of}t\mathrm{and}p\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ \frac{{\left(1+\log x\right)}^2}{4}=\frac{\log \left(1-y^2\right)}{2}+C.....\left(3\right) \\ \mathrm{At}x=1\mathrm{and}y=0,\left(3\right)\mathrm{becomes} \\ C=\frac{1}{4} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(3\right),\mathrm{we}\mathrm{get} \\ \frac{{\left(1+\log x\right)}^2}{4}=\frac{\log \left(1-y^2\right)}{2}+\frac{1}{4} \\ \Rightarrow {\left(1+\log x\right)}^2=2\log \left(1-y^2\right)+1 \\ \mathrm{Or} \\ {\left(\log x\right)}^2+\log x^2=\log {\left(1-y^2\right)}^2 \\ \mathrm{It}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{particular}\mathrm{solution}. \end{gathered}$$