Question

The solution of the differential equation
$y\sqrt{1-x^2}dy+x\sqrt{1-y^2}dx=0$ is

• A. $\sqrt{1-x^2}=c$
• B. $\sqrt{1-y^2}=c$
• C. $\sqrt{1-x^2}+\sqrt{1-y^2}=c$
• D. $\sqrt{1+x^2}+\sqrt{1+y^2}=c$

Answer 1

The correct option is C $\sqrt{1-x^2}+\sqrt{1-y^2}=c$

$y\sqrt{1-x^2}dy+x\sqrt{1-y^2}dx=0$
Using variable separable,
$\frac{y.dy}{\sqrt{1-y^2}}=\frac{-x.dx}{\sqrt{1-x^2}}$
On integrating, $\int \frac{ydy}{\sqrt{1-y^2}}=-\int \frac{xdx}{\sqrt{1-x^2}}$
Put $1-y^2=u^2$ and $1-x^2=v^2$
i.e., $ydy=-udu$ and $xdx=-vdv$
so, $\int -\frac{udu}{u}=-\int -\frac{vdv}{v}$
$\Rightarrow -u=v+c$
$-\sqrt{1-y^2}=\sqrt{1-x^2}+c$
or $\sqrt{1-x^2}+\sqrt{1-y^2}=c$