A) \[\tan x\]
B) \[x(x-\pi )\]
C) \[(x-\pi )(1-{{e}^{x}})\]
D) \[{{\sec }^{2}}x\]
Correct Answer: A
Solution :
[a] Given \[\frac{dy}{dx}=1+{{y}^{2}}\] \[\Rightarrow \frac{dy}{1+{{y}^{2}}}=dx\Rightarrow {{\tan }^{-1}}y=x+c\] \[\Rightarrow y=\tan (x+c)\], now \[y(0)=0\Rightarrow tanc=0\] \[y(\pi )=0\Rightarrow tan(\pi +c)=0\Rightarrow c=n\pi \] \[\therefore y=\tan x\]You need to login to perform this action.
You will be redirected in
3 sec