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