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प्रश्न
Solve `("d"y)/("d"x) = (x + y + 1)/(x + y - 1)` when x = `2/3`, y = `1/3`
उत्तर
`("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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