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प्रश्न
Solve `x^2 "dy"/"dx" - xy = 1 + cos(y/x)`, x ≠ 0 and x = 1, y = `pi/2`
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उत्तर
Given equation can be written as
`x^2 "dy"/"dx" - xy = 2cos^2 (y/2x)`, x ≠ 0.
⇒ `(x^2 "dy"/"dx" - xy)/(2cos^2 (y/(2x))` = 1
⇒ `sec^2 (y/(2x))/2 [x^2 "dy"/"dx" - xy]` = 1
Dividing both sides by x3, we get
`sec^2(y/(2x))/2 [(x "dy"/"dx" - y)/x^2] = 1/x^3`
⇒ `"d"/"dx"[tan(y/(2x))] = 1/x^3`
Integrating both sides, we get
`tan(y/(2x)) = (-1)/(2x^2) + "k"`
Substituting x = 1, y = `pi/2`, we get
k = `3/2`
Therefore, `tan(y/(2x)) = -1/(2x^2) + 3/2` is the required solution.
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