Question

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

• A. $x+y+C$
• B. $x+2y+C$
• C. $x–y+C$
• D. $2x+y+C$

Answer 1

The correct option is C

$x–y+C$

Finding the value :
It is given that

$dx+dy=(x+y)(dx-dy)$

Hence,

$\begin{array}{rcl}-(x+y-1)dx& =& -(x+y+1)dy\\ \frac{dy}{dx}& =& \frac{x+y-1}{x+y+1}\\ Substitutingx+y& =& v\\ 1+\frac{dy}{dx}& =& \frac{dv}{dx}\\ \frac{dv}{dx}-1& =& \frac{v-1}{v+1}\\ \frac{dv}{dx}& =& \frac{2v}{v+1}\\ \frac{2v}{v+1}dx& =& dv\\ \frac{v+1}{2v}dv& =& dx\\ & & Integratingbothside,\\ \int \frac{v+1}{2v}dv& =& \int dx\\ \frac{1}{2}\left(v+\log v\right)& =& x+c\\ x+y+\log (x+y)& =& 2x+c\\ \log (x+y)& =& x-y+c\end{array}$

Hence, the correct answer is option C