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प्रश्न
For the differential equation, find the general solution:
`x dy/dx + y - x + xy cot x = 0(x != 0)`
उत्तर
Given differential equation
`x dy/dx + y - x + xy cot x = 0`
⇒ `x dy/dx + y (1 + x cot x) = x`
or `dy/dx + (1/x + cot x) y = 1` ...(i)
Comparing with `dy/dx + Py = Q`
`P = 1/x + cot x` and Q = 1
∴ `I.F. = e^(int P dx) = e^(int(1/x + cot x)dx)`
`= e^(log x) + log sin x`
`=> e^(log (x sin x)) = x sin x`
Hence the required solution
∴ `y × I.F. = int I.F. xx Q dx + C`
`=> y xx x sin x = int 1 * x sin x dx + C`
`=> xy sin x = - x cos x + int 1 cos x dx + C`
`=> xy sin x = - x cos x + sin x + C`
⇒ y = `(- x cos x)/(x sin x) + (sin x)/(x sin x) + C/(x sin x)`
⇒ `y = 1/x - cot x + C/ (x sin x)`
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