Question

If $dx+dy=(x+y)(dx-dy)$ then fine the value of log(x+y).

• A. $x+y+c$
• B. $x+2y+c$
• C. $x-y+c$
• D. $2x+y+c$

Answer 1

The correct option is C $x-y+c$

$dx+dy=xdx+ydx-xdy-ydy$

$1+\frac{dy}{dx}=(x+y)-x\frac{dy}{dx}-y\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{x+y-1}{x+y+1}$

Let $x+y=v$

$1+\frac{dy}{dx}=\frac{dv}{dx}\Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1$

$\Rightarrow \frac{dv}{dx}-1=\frac{v-1}{v+1}$

$\Rightarrow \frac{dv}{dx}=\frac{2v}{v+1}$

$|\frac{1}{2}[1+{}^1{/}_v]dv=|dx+e$

$\Rightarrow \frac{1}{2}[v+logv]=x+c$

$\Rightarrow log(x+y)=x-y+c$