A) \[x+{{e}^{y}}+{{({{\tan }^{-1}}x)}^{2}}=C\]
B) \[x+{{e}^{-y}}+{{({{\tan }^{-1}}x)}^{2}}=C\]
C) \[{{e}^{-y}}\left( x+{{({{\tan }^{-1}}x)}^{2}} \right)=C\]
D) \[{{e}^{y}}\left( x+{{({{\tan }^{-1}}x)}^{2}} \right)=C\]
Correct Answer: A
Solution :
[a] The given differential equation is \[\left( \frac{2{{\tan }^{-1}}x}{1+{{x}^{2}}}+1 \right)dx+{{e}^{y}}dy=0\] \[\Rightarrow \,\,\,\,{{({{\tan }^{-1}}x)}^{2}}+x+{{e}^{y}}=C\]
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