A) \[6(x+y)-11\text{ }log\text{ (}3x+3y+10)=9x+c\]
B) \[6(x+y)+11\text{ }log\text{ (}3x+3y+10)=9x+c\]
C) \[6(x+y)-\log \left( x+y+\frac{10}{3} \right)=9x+c\]
D) \[6(x+y)\left( x+y+\frac{10}{3} \right)=9x+c\]
Correct Answer: A
Solution :
Given, \[\frac{dy}{dx}=\frac{x+y+7}{2x+2y+3}\] ...(i) Let \[x+y=v\] \[\Rightarrow \] \[\frac{dy}{dx}+1=\frac{dv}{dx}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{dv}{dx}-1\] On putting this value in Eq. (i), we get \[\frac{dv}{dx}-1=\frac{v+7}{2v+3}\] \[\Rightarrow \] \[\frac{dv}{dx}=\frac{v+7}{2v+3}\] \[\Rightarrow \] \[\frac{dv}{dx}=\frac{v+7+2v+3}{2v+3}\] \[\Rightarrow \] \[\frac{dv}{dx}=\frac{3v+10}{2v+3}\] \[\Rightarrow \] \[\frac{2v+3}{3v+10}dv=dx\] \[\Rightarrow \] \[\left( \frac{2}{3}-\frac{11}{3(3v+10)}dv \right)=dx\] \[\Rightarrow \] \[\frac{2}{3}dv-\frac{11}{3(3v+10)}dv=dx\] On integrating, we get \[\frac{2}{3}v=-\frac{11}{3}\log (3v+10).\frac{1}{3}+{{c}_{1}}=x\] \[\Rightarrow \]\[6v-11\log (3v+10)9{{c}_{1}}=9x\] \[\Rightarrow \]\[6(x+y)-11\log \{3(x+y)+10\}=9x-9{{c}_{1}}\] \[\Rightarrow \]\[6(x+y)-11\log (3x+3y+10)=9x-c\] where \[c=-9{{c}_{1}}\]
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