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महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

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 ∴ sec2xtanx dx+□ = 0 Integrating, we get □+∫sec2ytany dy = log c Each

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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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पाठ 1.8: Differential Equation and Applications - Q.6

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