Q. 414.8( 4 Votes )

Solution of the differential equation t any sec2x dx + tanx sec2 ydy = 0 is:
A. tanx + tany = k

B. tanx – tan y = k

C.

D. tanx . tany = k

Answer :

The given differential equation is


tanysec2xdx + tanxsec2ydy = 0


Divide throughout by tanxtany



Integrate



Put tanx = t hence that is sec2xdx = dt


Put tany = z hence that is sec2ydy = dt



log t + log z + c = 0


Resubstitute t and z


log(tan x) + log(tan y) + c = 0


Using log a + log b = log ab


log(tan x tan y) = -c


tan x tan y = e-c


e is a constant -c is a constant hence e-c is a constant, let it be denoted as k hence k = e-c


tan x tan y = k

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