Q. 264.5( 2 Votes )

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

Answer :

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



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