To solve the given differential equation:
(sin‌ycos2y−x sec2y)dy=(tan‌y)‌dx,we start by rewriting it in the standard linear form:
‌+‌=cos3y.Recognizing this as a linear differential equation in
x, we determine the integrating fac
Compute the Integrating Factor (IF):
IF=e∫‌dy.Solve the Integral:
∫‌dy=2‌∫csc‌2‌y‌d‌y.Solving this integral gives:
e∫csc‌2‌y‌d‌y=elog|csc‌2‌y−cot‌2‌y|=csc‌2‌y−cot‌2‌y.Thus, the integrating factor simplifies to:
IF=‌=‌| 2sin‌2y |
| 2sin‌y‌cos‌y |
=tan‌y.Apply the Integrating Factor:
Multiply the entire linear differential equation by this integrating factor:
x×tan‌y=∫cos3y×tan‌y‌d‌yIntegrate the Right-Hand Side:
Substitute
cos2y=1−sin‌2y. Then:
∫cos3y‌tan‌y‌d‌y=∫cos2ysin‌ydy.Perform the integration:
=−‌+C.Resulting General Solution:
Therefore, the general solution to the differential equation is:
3x‌tan‌y+cos3y=C.