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प्रश्न
Find the particular solution of the differential equation `dy/dx` = e2y cos x, when x = `π/6`, y = 0
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उत्तर
`dy/dx` = e2y cos x
∴ `dy/e^(2y)` = cos x.dx
∴ e–2y.dy = cos x.dx
Integrating both sides, we get
∴ `inte^(-2y).dy = int cos x.dx`
∴ `e^(-2y)/(-2)` = sin x + c ...(I)
When x = `π/6`, y = 0.
So equation (1), becomes
∴ `e^0/(-2) = sin π/6 + c`
∴ `-1/2 = 1/2 + c`
∴ `-1/2 - 1/2` = c ...(Given initial condition determines the value of c)
∴ c = –1
Put in equation (1), we get
∴ `e^(-2y)/(-2)` = sin x – 1
∴ e–2y = 2sin x – 2
∴ e2y (2sin x – 2) + 1 = 0 is the required particular solution.
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