Question

Solve:

`2(y + 3) - xy  (dy)/(dx)` = 0, given that y(1) = – 2.

Solve the differential equation `2(y + 3) - xy "dy"/"dx"` = 0, given that y(1) = – 2.

Answer 1

Given differential equation is `2(y + 3) - xy "dy"/"dx"` = 0

⇒ `xy  "dy"/"dx"` = 2y + 6

⇒ `(y/(2y + 6))"d"y = "dx"/x`

⇒ `1/2 (y/(y + 3))"d"y = "dx"/x`

Integrating both sides, we get

⇒ `1/2 int y/(y + 3)  dy = int dx/x`

⇒ `1/2 int (y - 3 - 3)/(y + 3)  dy = int dx/x`

⇒ `1/2 int (1 - 3/(y + 3)) dy = int (dx)/x`

⇒ `1/2 int dy - 3/2 int 1/(y + 3)  dy = int (dx)/x`

⇒ `1/2 y - 3/2 log |y + 3| = log x + c`

Put x = 1, y = –2

⇒ `1/2 (-2) - 3/2 log|-2 + 3| = log(1) + c`

⇒ `-1 - 3/2 log(1) = log(1) + c`

⇒ – 1 – 0 = 0 + c   ....[∵ log (1) = 0]

∴ c = –1

∴ Equation is `1/2 y - 3/2 log|y + 3| = log x - 1`

⇒ `y - 3 log |y + 3| = 2 log x - 2`

⇒ `y - 3 log|(y + 3)^3| = log x^2 - 2`

⇒ `log|(y + 3)^3| + log x^2 = y + 2`

⇒ `log|x^2 (y + 3)^3| = y + 2`

⇒ `x^2(y + 3)^3 = e^(y + 2)`

Hence, the required solution is `x^2(y + 3)^3 = e^(y + 2)`