Question

For the following differential equation find the particular solution satisfying the given condition:

3e^x tan y dx + (1 + e^x) sec² y dy = 0, when x = 0, y = π.

Answer 1

3e^x tan y dx + (1 + e^x) 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 + e^x| + log |tan y| = log c, where c₁ =log c

∴ log |(1 + e^x)³ * tan y| = log c 

∴ (1 + e^x)³ tan y = c

When x = 0, y = π, we have

(1 + e⁰)³ tan π = c

∴ c = 0

∴ the particular solution is (1 + e^x)³ tan y = 0