Question

The solution of the given D.E $\frac{dy}{dx}=\frac{y+\sqrt{x^2+y^2}}{x}$ 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

Answer: (A)

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

Put y=vx

$\frac{dy}{dx}=v=x\frac{dv}{dx}$

$\Rightarrow v+x\frac{dv}{dx}=v+\sqrt{1+v^2}$

$\Rightarrow x\frac{dv}{dx}=\sqrt{1+v^2}$

$\int \frac{dv}{\sqrt{1+v^2}}Putv=tan\theta$

$\Rightarrow \int \frac{\sec \theta d\theta }{\sec \theta }=\int \sec \theta d\theta =log(\sec \theta +\tan \theta )$

$\Rightarrow log(\sec \theta +\tan \theta )=cx$

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

$\Rightarrow \sqrt{1+v^2}+v=cx$

$\sqrt{x^2+y^2}+y=c{x}^2$ $[\therefore v=y/x]$