Q. 36 J5.0( 1 Vote )

Find one-parameter families of solution curves of the following differential equations:

e–ysec2ydy = dx + xdy

Answer :

e–ysec2ydy = dx + xdy


Given e–ysec2ydy = dx + xdy





This is a first order linear differential equation of the form



Here, P = 1 and e–ysec2y


The integrating factor (I.F) of this differential equation is,





We have


I.F = ey [ elog x = x]


Hence, the solution of the differential equation is,





Recall


xey = tan y + c



x = (tan y + c)e–y


Thus, the solution of the given differential equation is x = (tan y + c)e–y


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Interactive Quiz on DIfferential Calculus50 mins
Functional Equations - JEE with ease48 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses