# Find the general solution of (1 + tany) (dx – dy) + 2xdy = 0.

Given: (1 + tany) (dx – dy) + 2xdy = 0

dx – dy + tany dx – tany dy + 2xdy = 0

Divide throughout by dy

Divide by (1 + tany)

Compare with

we get and Q = 1

This is linear differential equation where P and Q are functions of y

For the solution of linear differential equation, we first need to find the integrating factor

IF = e∫Pdy

Put

Add and subtract siny in numerator

Consider the integral

Put siny + cosy = t hence differentiating with respect to y

we get which means

dt = (cosy – siny)dy

Resubstitute t

Hence the IF will be

IF = ey + log(siny+cosy)

IF = ey × elog(siny+cosy)

IF = ey(siny + cosy)

The solution of linear differential equation is given by x(IF) = ∫Q(IF)dy + c

Substituting values for Q and IF

xey(siny + cosy) = ∫(1)ey(siny + cosy)dy + c

xey(siny + cosy) = ∫(eysiny + eycosy)dy + c

Put eysiny = t and differentiating with respect to y we get which means dt = (eysiny + eycosy)dy

Hence

xey(siny + cosy) = ∫dt + c

xey(siny + cosy) = t + c

Resubstituting t

xey(siny + cosy) = eysiny + c

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
Interactive Quiz on Differential Calculus | Check Yourself56 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