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Solve the following differential equation
sec^{2} x tan y dx + sec^{2} y tan x dy = 0
Solution: sec^{2} x tan y dx + sec^{2} y tan x dy = 0
∴ `(sec^2x)/tanx "d"x + square` = 0
Integrating, we get
`square + int (sec^2y)/tany "d"y` = log c
Each of these integral is of the type
`int ("f'"(x))/("f"(x)) "d"x` = log f(x) + log c
∴ the general solution is
`square + log tan y` = log c
∴ log tan x . tan y = log c
`square`
This is the general solution.
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Solution
sec^{2} x tan y dx + sec^{2} y tan x dy = 0
∴ `(sec^2x)/tanx "d"x` + `(sec^2y)/tany "d"y` = 0
Integrating, we get
`int (sec^2x)/tanx "d"x` + `int (sec^2y)/tany "d"y` = log c
Each of these integral is of the type
`int ("f'"(x))/("f"(x)) "d"x` = log f(x) + log c
∴ the general solution is
log tan x + `log tan y` = log c
∴ log tan x . tan y = log c
∴ tan x . tan y = c
This is the general solution.
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