Differential Equations Question 76
Question: If $ y\cos x+x\cos y=\pi $ , then $ {{y}’}’(0) $ is
[IIT Screening 2005]
Options:
1
B) $ \pi $
0
D) $ -\pi $
Show Answer
Answer:
Correct Answer: B
Solution:
$ y\cos x+x\cos y=\pi $ Differentiate both sides with respect to x, we get $ -y\sin x+\cos x\cdot {y}’+x(-\sin y){y}’+\cos y $ Again differentiate with respect to x $ -{{y}’}’\sin x-y\cos x+\cos x\cdot {{y}’}’+\sin x\cdot {y}’-\sin y\cdot {y}’ $
$ -x[\cos y.{{({y}’)}^{2}}+\sin y.{{y}’}’]-\sin y.{y}’ $ Putting $ x=0 $ , we get $ -y+{{y}’}’-2\sin y{y}’=0 $
$ {{y}’}’=y+2{y}’\sin y $ Since at $ x=0 $ , $ y=\pi $ ; $ {{({{y}’}’)}_0}= -2\sin \pi $ .
BETA
