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Question
Choose the correct option from the given alternatives:
The solution of `("x + y")^2 "dy"/"dx" = 1` is
Options
x = tan-1 (x + y) + c
y tan-1 `("x"/"y") = "c"`
y = tan-1 (x + y) + c
y + tan-1 (x + y) + c
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Solution
y = tan-1 (x + y) + c
Hint:
`("x + y")^2 "dy"/"dx" = 1`
Put x + y = u ∴ `1 + "dy"/"dx" = "du"/"dx"`
∴ `"u"^2 ("du"/"dx" - 1) = 1`
∴ `"u"^2 "du"/"dx" = "u"^2 + 1`
∴ `int "u"^2/("u"^2 + 1) "du" = int "dx"`
∴ `int (("u"^2 + 1) - 1)/("u"^2 + 1) "du" = int "dx"`
∴ `int (1 - 1/"u")"du" = int "dx"`
∴ u - tan-1 u = x + c
∴ x + y - tan-1 (x + y) = x + c
∴ y = tan-1 (x + y) + c.
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