Solve dydx=x+y+1x+y-1 when x = 23, y = 13 - Mathematics and Statistics

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Sum

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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Chapter 1.8: Differential Equation and Applications - Q.4

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