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प्रश्न
For the following differential equation find the particular solution satisfying the given condition:
3ex tan y dx + (1 + ex) sec2 y dy = 0, when x = 0, y = π.
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उत्तर
3ex tan y dx + (1 + ex) sec2 y dy = 0, when x = 0, y = π.
∴`(3"e"^"x")/(1 + "e"^x) "dx" + ("sec"^2"y")/("tan y") "dy" = 0`
Integrating both sides, we get
`3 int "e"^"x"/(1 + "e"^"x") "dx" + int (sec^2"y")/(tan "y") "dy" = "c"_1`
Each of these integrals is of the type
`int ("f"'("x"))/("f"("x")) "dx" = log |"f"(x)| + "c"`
∴ the general solution is
3 log |1 + ex| + log |tan y| = log c, where c1 =log c
∴ log |(1 + ex)3 * tan y| = log c
∴ (1 + ex)3 tan y = c
When x = 0, y = π, we have
(1 + e0)3 tan π = c
∴ c = 0
∴ the particular solution is (1 + ex)3 tan y = 0
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