JEE Main & Advanced
Mathematics
Differential Equations
Question Bank
Critical Thinking
question_answer
If \[y\cos x+x\cos y=\pi \], then \[{{y}'}'(0)\] is [IIT Screening 2005]
A)1
B)\[\pi \]
C)0
D)\[-\pi \]
Correct Answer:
B
Solution :
\[y\cos x+x\cos y=\pi \] Differentiate both sides with respect to x, we get \[-y\sin x+\cos x.{y}'+x(-\sin y){y}'+\cos y\] Again differentiate with respect to x \[-{{y}'}'\sin x-y\cos x+\cos x.{{y}'}'+\sin x.{y}'-\sin y.{y}'\]
\[-x[\cos y.{{({y}')}^{2}}+\sin y.{{y}'}']-\sin y.{y}'\] Putting \[x=0\], we get \[-y+{{y}'}'-2\sin y\,{y}'=0\] \[{{y}'}'=y+2{y}'\sin y\] Since at \[x=0\], \[y=\pi \]; \[{{({{y}'}')}_{0}}=\pi \].