# 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 :

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