CGPET 2013 Solved Paper

Given, differential equation is
‌cos(‌

dy

dx

)=a
‌⇒‌‌‌

dy

dx

=cos−1a
‌⇒‌‌dy=cos−1‌ adx ‌
‌
On integrating both sides, we get
∫dy‌=cos−1a‌∫dx+C
⇒‌‌y‌=cos−1ax+C . . . (i)
When ‌‌x=0, then y=2
Then, from Eq. (i), we get
2=0+C⇒C=2
On putting the value of C in Eq. (i), we get
y‌=xcos−1a+2
⇒‌‌‌y−2‌=xcos−1a
⇒‌‌‌

y−2

x

‌=cos−1a
⇒cos(‌

y−2

x

)‌=a
which is the required solution.