Differential Equations Question 358
A continuously differentiable function $ \varphi (x) \in (0,\pi ) $ satisfying $ y’ = 1 + y^{2},\ y(0) = 0 = y(\pi ) $ is
[MP PET 2000]
Options:
A) $ \tan x $
B) $ x(x-\pi ) $
C) $ (x-\pi ) $
$ (1-e^{x}) $
D) Not possible
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{dy}{dx}=1+y^{2} $
Therefore $ \frac{dy}{1+y^{2}}=dx $
Integrating both sides,
$ \int{\frac{dy}{1+y^{2}}=\int{dx}} $
Therefore $ {{\tan }^{-1}}y=x+c $
At $ x=0 $,
$ y=0, $ then $ c=0 $
At $ x=\pi , $
$ y=0, $ then $ {{\tan }^{-1}}0=0 +c $
Therefore $ c=-\pi $
$ \therefore {{\tan }^{-1}}y=x $
Therefore $ y=\tan x=\varphi (x) $
Therefore, the solution is $ y=\tan x $
But $ \tan x $ is not continuous function in $ (0,\pi ) $
Hence, $ \varphi (x) $ is not defined in $ (0,\pi ) $ .
BETA
