Question

Find the particular solution of the differential equation
$(1-y^2)(1+\log x)dx+2xydy=0$, given that $y=0$ when $x=1$

Answer 1

Given, $(1-y^2)(1+\log x)dx+2xydy=0$

$\Rightarrow \frac{(1+\log x)dx}{x}=\frac{2ydy}{(1-y^2)}$

Integrating, we get $\int \frac{(1+\log x)dx}{x}=\int \frac{2ydy}{(1-y^2)}$

Let $\log x=t$, so $\frac{dx}{x}=dt$

Also, let $y^2=s$, so $2ydy=ds$

$\Rightarrow \int (1+t)dt=\int \frac{ds}{1-s}$

$\Rightarrow t+\frac{t^2}{2}=\ln (1-s)+c$

$\Rightarrow \log x+\frac{(\log x{)}^2}{2}=\ln (1-y^2)+c$

When $y=0,x=1$; we get $0+0=0+c$

$\Rightarrow c=0$

$\Rightarrow \log x+\frac{(\log x{)}^2}{2}=\ln (1-y^2)$