(√x + x) dy - (√y + y) dx = 0

(x+x)dy-(y+y)dx=0, 2dxy-2dxy=0

[tex]\displaystyle (\sqrt{x}+x)\,dy-(\sqrt{y}+y)\,dx=0\\(\sqrt{x}+x)\,dy=(\sqrt{y}+y)\,dx\\\frac1{\sqrt{y}+y}\,dy=\frac1{\sqrt{x}+x}\,dx\\\int\frac1{\sqrt{y}+y}\,dy=\int\frac1{\sqrt{x}+x}\,dx\\\\\text{misal :}\\\sqrt x=u\Rightarrow du= \frac1{2\sqrt{x}}\,dx=\frac1{2u}\,dx\\\sqrt y=v\Rightarrow dv= \frac1{2\sqrt{y}}\,dy=\frac1{2v}\,dy\\\\\int\frac{2v}{v+v^2}\,dv=\int\frac{2u}{u+u^2}\,du\\\int\frac{2}{1+v}\,dv=\int\frac{2}{1+u}\,du\\2\ln(v+1)=2\ln(u+1)\\\ln(\sqrt y+1)=\ln(sqrt x+1)[/tex]
[tex]\displaystyle \sqrt y=e^{\ln(\sqrt x+1)}-1\\\sqrt y=\sqrt x+1-1\\\sqrt y=\sqrt x\\\boxed{\boxed{y=x}}[/tex]