Differential Equations Question 141
Question: The solution of $ dy=\cos x(2-y\cos ecx)dx $ where $ y=2 $ when $ x=\frac{\pi }{2} $ is
[J & K 2005]
Options:
A) $ y=\sin x+\text{cosec }x $
B) $ y=\tan \frac{x}{2}+\cot \frac{x}{2} $
C) $ y=\frac{1}{\sqrt{2}}\sec \frac{x}{2}+\sqrt{2}\cos \frac{x}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{dy}{dx}=2\cos x-y\cot x $
Therefore $ \frac{dy}{dx}+y\cot x=2\cos x $ I.F. $ ={e^{\int{\cot xdx}}}=\sin x $
$ y.\sin x=\int{2\cos x.\sin x+c} $
$ y\sin x={{\sin }^{2}}x+c $ at $ y=2 $ and $ x=\frac{\pi }{2} $ , $ c=1 $ ; $ y=\sin x+\cos ecx $
BETA
