Question

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

• A. $sin{h}^{-1}+sin{h}^{-1}y=c$
• 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

Answer: (B)

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

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

let $t=\sqrt{1+x^2}$ ;$k=\sqrt{1+y^2}$

$\Rightarrow dt=\frac{x}{\sqrt{1+x^2}}dx$ ;$dk=\frac{y}{\sqrt{1+y^2}}dy$

$\Rightarrow \int dt+\int dk=0$

$\Rightarrow t+k=c$

Hence, Option B is correct