Solve the differential equation : `("x"^2 + 3"xy" + "y"^2)d"x" - "x"^2 d"y" = 0 "given that" "y" = 0 "when" "x" = 1`.

#### Solution

Here `("x"^2 + 3"xy" + "y"^2)d"x" - "x"^2d"y" = 0`

⇒ `(d"y")/(d"x") = ("x"^2 + 3"xy"+"y"^2)/("x"^2)`

⇒ `(d"y")/(d"x") = 1 + 3("y")/("x") + ("y"^2)/("x"^2)`

Put y = vx

⇒ `(d"y")/(d"x") = "v" + "x"(d"v")/(d"x")`

∴ `"v" + "x"(d"v")/(d"x") = 1 + 3"v"+"v"^2`

⇒ `"x"(d"v")/(d"x") = 1 + 2"v"+"v"^2`

⇒ `int_ (d"v")/(("v"+1)^2) = int_ (d"x")/("x")`

⇒ `-(1)/(("v"+1)) = log|"x"| + "C"`

⇒ `-("x")/("y"+"x") = log|"x"|+"C"`

As 1 y = 0 when x = 1 so, `-(1)/(0+1) = log|1|+"C"`

⇒ `"C" = -1`.

Hence the required solution is, `-("x")/("y"+"x") = log|"x"|-1`

⇒ `-"x" = "y" log|"x"| +"x" log |"x"| -"y" -"x"`

∴ `"y" = log|"x"| ("x" + "y")`

or,

`"y" = ("x" log|"x"|)/(1-log|"x"|)`.