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Question
For the differential equation find a particular solution satisfying the given condition:
`dy/dx` = y tan x; y = 1 when x = 0
Solution
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.
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