Answer:
We have, \[xy\frac{dy}{dx}=(x+2)\,(y+2)\Rightarrow \frac{y}{y+2}dy=\frac{x+2}{x}\,dx\] \[\Rightarrow \] \[\left( 1-\frac{2}{y+2} \right)\,dy=\left( 1+\frac{2}{x} \right)\,\,dx\] On integrating, we get \[y-2\,\,\log \,\,(y+2)=x+2\,\,\log x+C\]?(i) Given,\[y=-\,1,\] when \[x=1,\]then from Eq. (i), \[-\,1-2\,\log \,(-\,1+2)=1+2\,\log 1+C\] \[[\because \log 1=0]\] \[\Rightarrow \] \[C=-\,2,\] Then eq. (i) becomes \[y-2\,\,\log \,\,(y+2)=x+2\,\,\log \,\,x-2\] \[\Rightarrow \] \[y=x-2+2\,\{\log \,(y+2)+\log x\}\] \[\Rightarrow \] \[y=x-2+2\,\,\log \,\,\{x\,(y+2)\}\] Which is the required particular solution. OR We have, \[x\,({{x}^{2}}-1)\frac{dy}{dx}=1\Rightarrow dy=\frac{1}{x\,({{x}^{2}}-1)}\,dx\] Put \[({{x}^{2}}-1)=t\Rightarrow 2xdx=dt\] Now,\[\int{dy=\frac{1}{2}}\int{\frac{dt}{t\,(t+1)}}\] Resolving into partial fractions, we get \[\int{dy=\frac{1}{2}}\left[ \int{\,\left( \frac{1}{t}-\frac{1}{t+1} \right)\,dt} \right]\] \[\Rightarrow \] \[y=\frac{1}{2}[\log t-\log |t+1|]+C\] \[\Rightarrow \] \[y=\frac{1}{2}[\log t-\log [t+1]]+C\] \[\Rightarrow \] \[y=\frac{1}{2}\log \left| \frac{t}{t+1} \right|+C\] ?(i) \[\Rightarrow \] \[y=\frac{1}{2}\log \left| \frac{{{x}^{2}}-1}{{{x}^{2}}} \right|\] When \[x=2\]then\[y=0\] \[\therefore \] \[C=-\frac{1}{2}\log \frac{3}{4}\] From Eq. (i), \[y=\frac{1}{2}\log \,\left( 1-\frac{1}{{{x}^{2}}} \right)-\frac{1}{2}\log \frac{3}{4}\] which is the required particular solution.
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