Question

The solution of $x\sqrt{1+y^2}dx+y\sqrt{1+x^2}dy=0$.

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

Answer 1

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

Given,

$x\sqrt{1+y^2}dx+y\sqrt{1+x^2}dy=0$

separating the variable,

$x\sqrt{1+y^2}dx=-\sqrt{1+x^2}dy$

$=\frac{x}{\sqrt{1+x^2}}dx=-\frac{y}{\sqrt{1+y^2}}dy$

Let $1+x^2=t$

So, $2xdx=dt$

Let $1+y^2=u$

$2ydy=du$

substituting,

$\frac{dt}{2\sqrt{t}}=-\frac{dt}{2\sqrt{u}}$

Integrating,

$\frac{1}{2}\int \frac{dt}{\sqrt{t}}=\frac{-1}{2}\int \frac{du}{\sqrt{t^2}}$

$\frac{1}{2}\int t-\frac{1}{2}dt=\frac{-1}{2}\int -\frac{1}{2}\int u-\frac{1}{2}du$

as $\int x^{n}dx=\frac{x^{n+1}}{n+1}$

$=\frac{1}{2}\frac{t^{-1/2+1}}{\frac{-1}{2}+1}=\frac{-1}{2}\frac{u^{-1/2+1}}{\frac{-1}{2}+1}+c$

$=\frac{t^{1/2}}{1/2}=\frac{-u^{1/2}}{1/2}+c$

$=t^{1/2}+u^{1/2}=c$

$=\sqrt{t}+\sqrt{u}=c$

$=\sqrt{1+x^2}+\sqrt{1+y^2}=c$

So, the solution of $x\sqrt{1+y^2}dx+y\sqrt{1+x^2}dy=0$ is $\sqrt{1+x^2}+\sqrt{1+y^2}=c$