If $\quad d x+d y=(x+y)(d x-d y), \quad$ then $\log (x+y)$ is equal to

• A

  $x+y+c$
• B

  $x+2y+c$
• Correct answer
  C

  $x-y+c$
• D

  $2 x+y+c$

Solution

Given that $d x+d y=(x+y)(d x-d y)$ $\Rightarrow \quad d x(x+y-1)=d y(x+y+1)$
Let $x+y=t$ $\Rightarrow \quad \frac{d y}{d x}=\frac{d t}{d x}-1$ $\therefore$ From Eq. (i) $\begin{aligned} \frac{d t}{d x}-1 & =\frac{t-1}{t+1} \\ \Rightarrow \quad \frac{d t}{d x} & =\frac{t-1+t+1}{t+1} \\ \Rightarrow \quad \frac{(t+1)}{2 t} d t & =d x \\ \Rightarrow \quad \frac{1}{2}\left(1+\frac{1}{t}\right) d t & =d x \end{aligned}$ On integrating both sides, we get $\begin{aligned} \frac{1}{2}(t+\log t) & =x+\frac{C}{2} \\ \log (x+y) & =2 x-x+y+C \\ \log (x+y) & =x-y+C \end{aligned}$

Hence, the correct answer is option: 'C'.