Question

The solution of $\frac{dy}{dx}=\frac{x+y}{x-y}$ is

• A. ${\tan }^{-1}\left(\frac{y}{x}\right)=\log \sqrt{x^2-y^2}+C$
• B. ${\tan }^{-1}\left(\frac{y}{x}\right)=\log \sqrt{x^2-y^2}+C$
• C. ${\sin }^{-1}\left(\frac{y}{x}\right)=\log \sqrt{x^2+y^2}+C$
• D. ${\cos }^{-1}\left(\frac{y}{x}\right)=\log \sqrt{x^2-y^2}+C$

Answer 1

Answer: (A)

We have,
$\frac{dy}{dx}=\frac{x+y}{x-y}$
Put $y=vx$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$
$\therefore v+x\frac{dv}{dx}=\frac{x+vx}{x-vx}=\frac{1+v}{1-v}$
$\Rightarrow x\frac{dv}{dx}=\frac{1+v}{1-v}-v=\frac{1+v-v+v^2}{1-v}=\frac{1+v^2}{1-v}$
$\Rightarrow \frac{1-v}{1+v^2}dv=\frac{dx}{x}$
$\Rightarrow \int \frac{1}{1+v^2}dv-\frac{1}{2}\int \frac{2v}{1+v^2}dv=\int \frac{dx}{x}$
$\Rightarrow {\tan }^{-1}v-\frac{1}{2}\log \left|1+v^2\right|=\log x+C$
$\Rightarrow {\tan }^{-1}\frac{y}{x}=\log x+\frac{1}{2}\log \left(1+\frac{y^2}{x^2}\right)+C$
$\Rightarrow {\tan }^{-1}\frac{y}{x}=\log \frac{x\left(\sqrt{x^2+y^2}\right)}{x}+C$
$\Rightarrow {\tan }^{-1}\frac{y}{x}=\log \sqrt{x^2+y^2}+C$