Question

Solve the differential equation $x(1+y^2)dx-y(1+x^2)dy=0$

Answer 1

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

$\therefore \frac{y}{1+y^2}dy=\frac{x}{1+x^2}dx$ (variable separable)

Integrating both sides we get

$\int \frac{y}{1+y^2}dy=\int \frac{x}{1+x^2}dx\to 1$

$\to I_1=\int \frac{y}{1+y^2}dy$

$1+y^2=4\Rightarrow 2ydy=du$

$\therefore ydy=1\frac{du}{2}$

$\therefore I_1=\int \frac{1}{2}\frac{du}{u}$

$\therefore I_1=\frac{1}{2}\left(\log \left|u\right|\right)+c_1=\frac{1}{2}\log |1+y^2|+c_1$

$I_2=\int \frac{x}{1+x^2}dx$

$1+x^2=v$

$\therefore 2xdx=dv$

$\therefore xdx=dv/2$

$\therefore I_2=\frac{1}{2}\int \frac{dv}{v}=\frac{1}{2}\log \left(v\right)+c_2$

$=\frac{1}{2}\log |1+x^2|+c_2$

Now from $1$

$\frac{1}{2}\log |1+y^2|+c_1=\frac{1}{2}\log |1+x^2|+c_2$

$\therefore \frac{1}{2}\log |1+y^2|=\frac{1}{2}\log |1+x^2|+c$

where $c=c_2-c_1$

Put $x=1$ & $y=0$ in above equation

$\therefore c=\frac{-1}{2}\log 2$

$\therefore \frac{1}{2}\log |1+y^2|=\frac{1}{2}\log |1+x^2|-\frac{1}{2}\log 2$

$\therefore \log |1+y^2|=\log \left|\frac{1+x^2}{2}\right|$

$\therefore 1+y^2=\frac{1+x^2}{2}$

$\therefore 1+x^2=2(1+y^2)$