Question

Find the particular solution of the differntials equation $x(1+y^2)dx-y(1+x^2)dy=0$, given that $y=1$ when $x=0$.

Answer 1

$\text{Given differeential equation is}$
$x(1+y^2)dx-y(1+x^2)dy=0$

$\text{Separating the variables}$

$\Rightarrow \frac{x}{1+x^2}dx-\frac{y}{1+y^2}dy=0$,
$\text{Integrating both sides}$

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

$\Rightarrow log\frac{1+x^2}{1+y^2}=log{k}^2$

$\Rightarrow \frac{1+x^2}{1+y^2}=k^2,$$\text{when x=0, y=1}$

$\Rightarrow \frac{1}{2}=k^2$

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

$\Rightarrow y^2=1+2x^2$