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Solve `("d"y)/("d"x) = (x + y + 1)/(x + y  1)` when x = `2/3`, y = `1/3`
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Solution
`("d"y)/("d"x) = (x + y + 1)/(x + y  1)` ......(i)
Put x + y = u ......(ii)
∴ y = u − x
Differentiating w.r.t. x, we get
`("d"y)/("d"x) = ("du")/("d"x)  1` .....(iii)
Substituting (ii) and (iii) in (i), we get
`("du")/("d"x)  1 = ("u" + 1)/("u"  1)`
∴ `("du")/("d"x) = ("u" + 1)/("u"  1) + 1`
= `("u" + 1 + "u"  1)/("u"  1)`
∴ `("du")/("d"x) = (2"u")/("u"  1)`
∴ `(("u"  1)/"u") "du"` = 2dx
∴ `(1  1/"u") "du"` 2dx
Integrating on both sides, we get
`int(1  1/"u") "du" = 2int "d"x`
∴ u − log u = 2x + c
∴ x + y − log x + y = 2x + c
∴ − log x + y = x − y + c
Putting x = `2/3` and y = `1/3`, we get
− log (1) = `1/3 + "c"`
∴ c = `1/3`
∴ − log x + y = `x  y  1/3`
∴ log x + y = `y  x + 1/3`
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