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प्रश्न
Solve the following differential equation
sec2 x tan y dx + sec2 y tan x dy = 0
Solution: sec2 x tan y dx + sec2 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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उत्तर
sec2 x tan y dx + sec2 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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