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प्रश्न
Solve the following differential equation:
`2"e"^("x + 2y") "dx" - 3"dy" = 0`
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उत्तर
`2"e"^("x + 2y") "dx" - 3"dy" = 0`
∴ `2"e"^"x" * "e"^"2y" "dx" - 3"dy" = 0`
∴ `2"e"^"x" "dx" - 3/"e"^"2y" "dy" = 0`
Integrating both sides, we get
`2 int "e"^"x" "dx" - 3 int "e"^(-2"y") "dy" = "c"_1`
∴ `2"e"^"x" - 3 * ("e"^(- "2y"))/(- 2) = "c"_1`
∴ `4"e"^"x" + 3"e"^(- 2"y") = 2"c"_1`
∴ `4"e"^"x" + 3"e"^(- 2"y") = "c"`, where c = 2c1.
This is the general solution.
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