AP EAPCET 18 May 2024 Shift 1 Solved Paper

To solve the given differential equation:
(sin‌ycos2y−x sec2y)dy=(tan‌y)‌dx,
we start by rewriting it in the standard linear form:
‌

dx

dy

+‌

x

sin‌y‌cos‌y

=cos3y.
Recognizing this as a linear differential equation in x, we determine the integrating fac
Compute the Integrating Factor (IF):
IF=e∫‌

1

sin‌y‌cos‌y

dy.
Solve the Integral:
∫‌

1

sin‌y‌cos‌y

dy=2‌∫csc‌2‌y‌d‌y.
Solving this integral gives:
e∫csc‌2‌y‌d‌y=elog|csc‌2‌y−cot‌2‌y|=csc‌2‌y−cot‌2‌y.
Thus, the integrating factor simplifies to:
IF=‌

1−cos‌2‌y

sin‌2y

=‌

2sin‌2y

2sin‌y‌cos‌y

=tan‌y.
Apply the Integrating Factor:
Multiply the entire linear differential equation by this integrating factor:
x×tan‌y=∫cos3y×tan‌y‌d‌y
Integrate the Right-Hand Side:
Substitute cos2y=1−sin‌2y. Then:
∫cos3y‌tan‌y‌d‌y=∫cos2ysin‌ydy.
Perform the integration:
=−‌

cos3y

3

+C.
Resulting General Solution:
Therefore, the general solution to the differential equation is:
3x‌tan‌y+cos3y=C.