Question

Solve the following differential equation:
$\sqrt{1+x^2+y^2+x^2{y}^2}+xy\frac{dy}{dx}=0$

• A. (1+x^2)^1/2 + (1+y^2)^1/2 + 1/2log( ((1+x^2)^1/2 -1 /(1+x^2)^1/2 +1 )) =c
• B. (1+x^2)^1/2 + (1+y^2)^1/2 + 1/2 =c
• C. (1+x^2)+ (1+y^2) + 1/2log( ((1+x^2)^1/2 -1 /(1+x^2)^1/2 +1 )) =c
• D. (1+y^2)^1/2 + 2log( ((1+x^2)^1/2 -1 /(1+x^2)^1/2 +1 )) =c

Answer 1

The correct option is A (1+x^2)^1/2 + (1+y^2)^1/2 + 1/2log( ((1+x^2)^1/2 -1 /(1+x^2)^1/2 +1 )) =c

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

Squaring on both the sides, we get

$1+x^2+y^2+x^2{y}^2=x^2{y}^2(\frac{dy}{dx}{)}^2$

$1(1+x^2)+y^2(1+x^2)=x^2{y}^2(\frac{dy}{dx}{)}^2$

$\frac{(1+y^2)(1+x^2)}{x^2{y}^2}=(\frac{dy}{dx}{)}^2$

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

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

On LHS, we put $x=\tan \theta$

$dx={\sec }^2\theta d\theta$

$1+{\tan }^{\theta }={\sec }^2\theta$

$\Rightarrow \frac{\sec \theta }{\tan \theta }{\sec }^2\theta d\theta =\frac{y}{\sqrt{1+y^2}}dy$

$(\frac{1}{\sin \theta }\times \frac{1}{{\cos }^2\theta })d\theta =\frac{y}{\sqrt{1+y^2}}dy$

$\left(\frac{1+{\tan }^2\theta }{\sin \theta }\right)d\theta =\frac{y}{\sqrt{1+y^2}}dy$

$(\csc \theta +\sec \theta \tan \theta )d\theta =\frac{y}{\sqrt{1+y^2}}dy$

Now, by integrating we get

$\ln \mid \csc \theta -\cot \theta \mid +\sec \theta +C=\frac{1}{2}(\sqrt{1+y^2}$

$(\ln \mid \csc \theta -\cot \theta \mid )+\sec \theta +C=\frac{1}{2}(\sqrt{1+y^2}$

as $x=\tan \theta ,\cot \theta =\frac{1}{x}$

$\csc \theta =\left(\frac{\sqrt{1+x^2}}{x}\right)$

$\sec \theta =\sqrt{1+x^2}$

$(\ln \mid \frac{\sqrt{1+x^2}}{x}-\frac{1}{x}\mid )+\sqrt{1+x^2}+C=\frac{1}{2}\sqrt{1+y^2}$