Differential Equations Question 241
Question: A continuously differentiable function $ \phi (x) $ , $ x\in [0,\pi ]-{ \frac{\pi }{2} } $ satisfying $ y’=1+y^{2},y(0)=0=y(\pi ) $ is
Options:
A) $ \tan x $
B) $ x(x-\pi ) $
C) $ (x-\pi )(1-e^{x}) $
D) $ {{\sec }^{2}}x $
Show Answer
Answer:
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 $
BETA
