A) \[x+y+\tan (x+c)=0\]
B) \[x-y+\tan (x+c)=0\]
C) \[x+y-\tan (x+c)=0\]
D) None of these
Correct Answer: C
Solution :
Put \[x+y=v\]and \[1+\frac{dy}{dx}=\frac{dv}{dx}\] Þ \[\frac{dv}{dx}={{v}^{2}}+1\]Þ\[\frac{dv}{{{v}^{2}}+1}=dx\] On integrating, we get \[{{\tan }^{-1}}v=x+c\]or \[v=\tan (x+c)\] Þ\[x+y=\tan (x+c)\].You need to login to perform this action.
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