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प्रश्न
Find a particular solution of the differential equation `dy/dx + y cot x = 4xcosec x(x != 0)`, given that y = 0 when `x = pi/2.`
उत्तर
Given the differential equation
`dy/dx + y cot x = 4x cosec x` ....(1)
Comparing with the linear equation `dy/dx + Py = Q`,
When P = cot x, Q = 4x cosec x
∴ `I.F. = e^(int Pdx) = e^(int cot x dx) = e^(log |sin x|) = sin x`
∴ The solution is `y. (I.F.) = int Q. (I.F.) dx + C`
`therefore y sin x = int 4x cosec x sin x dx + C`
`= int 4x dx + C = + C`
`= (4x^2)/2 + C`
⇒ y sinx = 2x2 + C ....(2)
When `x = pi/2, y = 0`
∴ `0 = 2 (pi^2/4) + C`
⇒ `C = -pi^2/2`
Putting `C = pi^2/2` in (2),
`y sinx = 2x^2 - pi^2/2 ; (sin x ne 0)`
Which is the required solution.
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