A) \[\tan \,x\]
B) \[x(x-\pi )\]
C) \[(x-\pi )\,(1-{{e}^{x}})\]
D) not possible
Correct Answer: A
Solution :
Given, \[y'=1+{{y}^{2}}\] \[\Rightarrow \] \[\frac{dy}{1+{{y}^{2}}}=dx\] \[\Rightarrow \] \[{{\tan }^{-1}}y=x+C\] (on integrating) At \[x=0,\,ta{{n}^{-1}}y=0+C\] \[\Rightarrow \] \[0=0+C\] \[\Rightarrow \] \[C=0\] \[\therefore \] \[{{\tan }^{-1}}y=x\] \[\Rightarrow \] \[y=\tan \,x\]You need to login to perform this action.
You will be redirected in
3 sec