Question

#JEE
$\tiny{ \red { \bold{\sqrt{1 + {x}^{2} } . \sqrt{1 + {y}^{2} } \: dx + x.y.dy = 0}}}$
Variable Separable ​

Answer 1

Step-by-step explanation:

We have,

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

$\rm \implies\sqrt{1 + {x}^{2} } . \sqrt{1 + {y}^{2} } \: dx = - x.y.dy$

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

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

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

$\rm\: In\:\: LHS\:\: put\:\: x=\tan(\theta)\: \implies\:dx=\sec^2(\theta)d\theta$

$\rm\: In\:\:RHS\:\: derivative \:\:of \:\: denominator\:\: present\:\: in\:\: numerator$

So,

$\rm \implies \int \frac{\sqrt{1 + \tan^{2} (\theta) }}{ \tan( \theta) } \: \sec ^{2} ( \theta) d \theta = - \frac{1}{2} . 2 \sqrt{1 + {y}^{2} } + c$

$\rm \implies \int \frac{ \sec (\theta) }{ \tan( \theta) } \: \sec ^{2} ( \theta) d \theta = -\sqrt{1 + {y}^{2} } + c$

$\rm \implies \int \frac{1 }{ \sin( \theta) } .\sec ^{2} ( \theta) d \theta = -\sqrt{1 + {y}^{2} } + c$

$\rm \implies \int \cosec( \theta) .\sec ^{2} ( \theta) d \theta = -\sqrt{1 + {y}^{2} } + c$

$\rm \implies \cosec( \theta) \int\sec ^{2} ( \theta) d \theta - \int \bigg \{ \frac{d}{d \theta} \{ \cosec( \theta) \} \int \sec^{2} ( \theta) d \theta \bigg \} d \theta= -\sqrt{1 + {y}^{2} } + c$

$\rm \implies \cosec( \theta) \tan( \theta) - \int \bigg \{ - \cosec( \theta) \cot( \theta) . \tan( \theta) \bigg \} d \theta= -\sqrt{1 + {y}^{2} } + c$

$\rm \implies \cosec( \theta) \tan( \theta) + \int \cosec( \theta) d \theta= -\sqrt{1 + {y}^{2} } + c$

$\rm \implies \cosec (\theta ) \tan( \theta) + ln \bigg | \tan \bigg( \frac{\theta}{2} \bigg) \bigg| = -\sqrt{1 + {y}^{2} } + c$

$\rm \implies \frac{ \sqrt{1 + {x}^{2} } }{x} . x + ln \bigg | \sqrt{ \frac{ \sqrt{1 + {x}^{2} } - 1}{ \sqrt{1 + {x}^{2} } + 1} } \bigg| = -\sqrt{1 + {y}^{2} } + c$

$\rm \implies \sqrt{1 + {x}^{2} } + \frac{1}{2} ln \bigg | \frac{ \sqrt{1 + {x}^{2} } - 1}{ \sqrt{1 + {x}^{2} } + 1} \bigg| = -\sqrt{1 + {y}^{2} } + c$