Question

The solution of the equation `1 + ("d"x)/("d"y) = (2y)/x` is ______.

पर्याय

• (x – y)(x + 2y)² = c

• y = x + c

• (x + y)(x – 2y)² = c

• y = `x/(2y - x) + "c"`

Answer 1

The solution of the equation `1 + ("d"x)/("d"y) = (2y)/x` is (x – y)(x + 2y)² = c.

Explanation:

 `1 + ("d"x)/("d"y) = (2y)/x` 

⇒ `("d"x)/("d"y) = (2y - x)/x`

⇒ `("d"y)/("d"x) = x/(2y - x)`  ......(i)

Put y = vx   ......(ii)

⇒ `("d"y)/("d"x) = "v" + x "dv"/("d"x)`  ......(iii)

Substituting (ii) and (iii) in (i), we get

`"v" + x "dv"/("d"x) = x/(2"v"x - x) = 1/(2"v" - 1)`

⇒ `x "dv"/("d"x) = 1/(2"v" - 1) = -"v" = (1 - 2"v"^2 + "v")/(2"v" - 1)`

⇒ `x "dv"/("d"x) = - (("v" - 1)(2"v" + 1))/(2"v" - 1)`

⇒ `((2"v" - 1))/((2"v" + 1)("v" - 1)) "dv" = (-"d"x)/x`

⇒ `1/(3("v" - 1)) + 4/(3(2"v" + 1)) = (-"d"x)/x`

Integrating on both sides, we get

`1/3 log("v" - 1) + 4/3, 1/2 log (2"v" + 1) = - logx + log"c"_1`

⇒ `log("v" - 1)^(1/3) + log(2"v" + 1)^(2/3) = log  "c"_1/x`

⇒ `("v" - 1)^(1/3) (2"v" + 1)^(2/3) = "c"_1/x`

⇒ `((y - x)/x) ((2y + x)/x)^3 = "c"_1^3/x^3`

⇒ (x – y)(x + 2y)² = c, where c = `-"c"_1^3`