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प्रश्न
Solve the following differential equation:
`log ("dy"/"dx") = 2"x" + 3"y"`
उत्तर
`log ("dy"/"dx") = 2"x" + 3"y"`
∴ `"dy"/"dx" = "e"^("2x" + "3y") = "e"^"2x"."e"^"3y"`
∴ `1/"e"^"3y" "dy" = "e"^"2x" "dx"`
Integrating both sides, we get
`int "e"^-"3y" "dy" = int "e"^"2x' "dx"`
∴ `int "e"^-3"y" "dy" = int "e"^"2x" "dx"`
∴ `("e"^(- "3y"))/-3 = "e"^"2x"/2 + "c"_1`
∴ `2"e"^-"3y" = - 3"e"^"2x" + 6"c"_1`
∴ `2"e"^-"3y" + 3"e"^"2x" = "c"`, where c = 6c1
This is the general solution.
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