Question

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

• A. $y+\sqrt{x^2+y^2}=c{x}^2$
• B. $x+\sqrt{x^2+y^2}=c{y}^2$
• C. $y+\sqrt{x^2+y^2}=cx$
• D. $x+\sqrt{x^2+y^2}=cy$

Answer 1

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

Let $y=x\tan \theta$
$\Rightarrow \frac{dy}{dx}=\tan \theta +x{\sec }^2\theta \frac{d\theta }{dx}$

$\Rightarrow \tan \theta +x{\sec }^2\theta \frac{d\theta }{dx}=\frac{x\tan \theta +\sqrt{x^2(1+{\tan }^2\theta )}}{x}$

$\Rightarrow \tan \theta +x{\sec }^2\theta \frac{d\theta }{dx}=\tan \theta +\sec \theta$

$\Rightarrow \int \sec \theta d\theta =\int \frac{dx}{x}$

$\Rightarrow \log |\sec \theta +\tan \theta |=\log x+\log c$

$\Rightarrow \log |\sec \theta +\tan \theta |=\log xc$

$\Rightarrow \sec \theta +\tan \theta =xc$

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

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

Hence, option A.