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प्रश्न
For the differential equation find a particular solution satisfying the given condition:
`[xsin^2(y/x - y)] dx + x dy = 0; y = pi/4 "when" x = 1`
उत्तर
The given equation is
`[x sin^2 (y/x) - y] dx + x dy = 0`
`dy/dx = (x sin^2 y/x - y)/x`
`= (y - x sin^2 y/x)/x`
`= y/x - sin^2 y/x` ....(1)
Put y = vx so that `dy/dx = v - sin^2 v`
∴ (1) become: v + x `(dv)/dx = v - sin^2 v`
⇒ `x (dv)/dx = - sin^2 v`
⇒ `cosec^2 v dv = - dx/x`
Integrating, `int cosec^2 v dv = - int dx/x`
⇒ - cot v = - log |x| + C
⇒ log |x| - cot v = C
⇒ `log |x| - cot y/x = C` ....(2)
When x = 1, `y = pi/4`
∴ `log |1| - cot pi/4 = C`
⇒ 0 - 1 = C
⇒ C = -1
Putting in (2), log |x| - cot `y/x` = -1
⇒ `cot (y/x) = log |x| + log |e|`
Which is the required solution.
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