Question

For the differential equation find a particular solution satisfying the given condition:

`dy/dx` = y tan x; y = 1 when x = 0

Answer 1

We have,

`dy/dx = y tan x`

⇒ `dy/y = tan x dx`                             ....(1)

Integrating (1) both sides, we get

⇒ `int dy/y = int tan x dx`

⇒ log y = log |sec x| + C

When x = 0, y = 1

⇒ log 1 = log |sec 0| + C

⇒ 0 = log 1 + C

⇒ C = 0

∴ log y = log |sec x|

Hence, the particular solution is y = sec x.