A) \[(4x+y+1)=\tan \,(2x+C)\]
B) \[{{(4x+y+1)}^{2}}=2\tan \,(2x+C)\]
C) \[{{(4x+y+1)}^{3}}=3\tan \,(2x+C)\]
D) \[(4x+y+1)=2\tan \,(2x+C)\]
Correct Answer: D
Solution :
\[\frac{dy}{dx}={{(4x+y+1)}^{2}}\] ?(i) Put \[4x+y+1=v\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{dv}{dx}-4\] \[\therefore \] From Eq. (1), \[\frac{dv}{dx}-4={{v}^{2}}\] \[\Rightarrow \] \[\frac{dv}{{{v}^{2}}+4}=dx\] \[\Rightarrow \] \[\frac{1}{2}{{\tan }^{-1}}\left( \frac{v}{2} \right)=x+C\] \[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{4x+y+1}{2} \right)=2x+C\] \[\Rightarrow \] \[4x+y+1=2\,\tan \,(2x+C)\]
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