A) \[{{\tan }^{-1}}\left( \frac{2y+1}{\sqrt{3}} \right)=x+{{x}^{2}}+c\]
B) \[4{{\tan }^{-1}}\left( \frac{2y+1}{\sqrt{3}} \right)=x+{{x}^{2}}+c\]
C) \[\sqrt{3}{{\tan }^{-1}}\left( \frac{3y+1}{3} \right)=4(1+x+{{x}^{2}})+c\]
D) \[{{\tan }^{-1}}\left( \frac{2y+1}{3} \right)=4(2x+{{x}^{2}})+c\]
E) \[4{{\tan }^{-1}}\left( \frac{2y+1}{\sqrt{3}} \right)=\sqrt{3}(2x+{{x}^{2}})+c\]
Correct Answer: E
Solution :
\[\frac{dy}{dx}=1+y+{{y}^{2}}+x(1+y+{{y}^{2}})\] \[\Rightarrow \] \[\frac{dy}{1+y+{{y}^{2}}}=(1+x)dx\] \[\Rightarrow \] \[\int{\frac{dy}{{{\left( y+\frac{1}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}}=\int{(1+x)}dx\] \[\Rightarrow \]\[\frac{1}{\frac{\sqrt{3}}{2}}{{\tan }^{-1}}\left( \frac{y+\frac{1}{2}}{\frac{\sqrt{3}}{3}} \right)=x+\frac{{{x}^{2}}}{2}+\frac{c}{2}\] \[\Rightarrow \]\[4{{\tan }^{-1}}\left( \frac{2y+1}{\sqrt{3}} \right)=\sqrt{3}(2x+{{x}^{2}})+c\]
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