Question

Evaluate :
$\frac{dy}{dx}=\frac{x+y+1}{2x+2y+3}$

Answer 1

$\frac{dy}{dx}=\frac{x+y+1}{2x+2y+3}$

Let $x+y=z$

$\Rightarrow 1+\frac{dy}{dx}=\frac{dz}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{dz}{dx}-1$

$\therefore \frac{dz}{dx}-1=\frac{z+1}{2z+3}\Rightarrow \frac{dz}{dx}=\frac{z+1}{2z+3}+1=\frac{3z+4}{2z+3}$

$\Rightarrow \left(\frac{2z+3}{3z+4}\right)dz=dx$

$\Rightarrow (\frac{2z}{3z+4}+\frac{3}{3z+4})dz=dx$

$\Rightarrow \{\frac{2}{3}\left(\frac{3z+4-4}{3z+4}\right)+\frac{3}{3z+4}\}dz=dx$

$\Rightarrow \frac{2}{3}dz+\frac{1}{3}\left(\frac{1}{3z+4}\right)dz=dx$

Integrating both sides we get

$\frac{2}{3}z+\frac{1}{9}log(3z+4)=x+c$

$\Rightarrow \frac{2}{3}(x+y)+\frac{1}{9}log(3x+3y+4)=x+c$

$\Rightarrow \frac{2}{3}y-\frac{1}{3}x+\frac{1}{9}log(3x+3y+4)=c$.