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 = e^{y + log(siny+cosy)}

⇒ IF = e^y × e^{log(siny+cosy)}

⇒ IF = e^y(siny + cosy)

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

Substituting values for Q and IF

⇒ xe^y(siny + cosy) = ∫(1)e^y(siny + cosy)dy + c

⇒ xe^y(siny + cosy) = ∫(e^ysiny + e^ycosy)dy + c

Put e^ysiny = t and differentiating with respect to y we get which means dt = (e^ysiny + e^ycosy)dy

Hence

⇒ xe^y(siny + cosy) = ∫dt + c

⇒ xe^y(siny + cosy) = t + c

Resubstituting t

⇒ xe^y(siny + cosy) = e^ysiny + c