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**Solve the following differential equation:**

`log ("dy"/"dx") = 2"x" + 3"y"`

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#### Solution

`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 = 6c_{1}

This is the general solution.

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