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If $\quad d x+d y=(x+y)(d x-d y), \quad$ then $\log (x+y)$ is equal to
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Verified Answer
The correct answer is:
$x-y+c$
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
$\begin{aligned}
\frac{1}{2}(t+\log t) & =x+\frac{C}{2} \\
\log (x+y) & =2 x-x+y+C \\
\log (x+y) & =x-y+C
\end{aligned}$
$\Rightarrow \quad d x(x+y-1)=d y(x+y+1)$
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
$\begin{aligned}
\frac{1}{2}(t+\log t) & =x+\frac{C}{2} \\
\log (x+y) & =2 x-x+y+C \\
\log (x+y) & =x-y+C
\end{aligned}$
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