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प्रश्न
For the differential equation given, find a particular solution satisfying the given condition:
`dy/dx - 3ycotx = sin 2x; y = 2` when `x = pi/2`
उत्तर
The given equation is
`dy/dx - 3 y cot x = sin 2x` ....(1)
Which is a linear equation of the type
`dy/dx + Py = Q`
Here P = - 3cot x and Q = sin 2x
∴ `intP dx = -3 int cot x dx = -3 log |sin x|`
∴ `I.F. = e^(-3log|sin x|)`
`= e^(log cosec^3 x)`
`= cosec^3 x`
∴ The solution is `y. (I.F.) = int Q. (I.F.) dx + C`
`y cosec^3 x = int sin2x cosec^3x dx + C`
`= int (2 sin x cos x)/(sin^3 x) dx + C`
`= 2 int cosec x cot x dx + C`
`= - 2 cosec x +C`
⇒ y = -2 sin2 x + C sin3 x ....(2)
When `x = pi/2, y = 2`
∴ `2 = -2 sin^2 pi/2 + C sin^3 pi/2`
⇒ 2 = -2 (1)2 + C (1)3
⇒ C = 2 + 2
⇒ C = 4
Putting in (2), we get
y = - 2sin2 x + 4 sin3 x
⇒ y = 4 sin3 x - 2 sin x
Which is the required solution.
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