Q72 / 170 · Manipal (MET) 2018 · Mathematics

The general solution of $x\left(1+y^2\right)^{1 / 2} d x+y\left(1+x^2\right)^{1 / 2} d y=0$ is

  • A
    $\sin ^{-1} x+\sin ^{-1} y=c$
  • B
    $x^2+y^2=\left(1+x^2\right)^{1 / 2}+\left(1+y^2\right)^{1 / 2}+c$
  • Correct answer
    C
    $\left(1+x^2\right)^{1 / 2}+\left(1+y^2\right)^{1 / 2}=c$
  • D
    $\tan ^{-1} x-\tan ^{-1} y=c$

Solution

Given equation can be rewritten as $\frac{x}{\left(1+x^2\right)^{1 / 2}} d x+\frac{y}{\left(1+y^2\right)^{1 / 2}} d y=0$
On integrating, we get $\sqrt{1+x^2}+\sqrt{1+y^2}=c$

Hence, the correct answer is option: 'C'.

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