Question

General solution of `(x + y)^2 ("d"y)/("d"x) = "a"^2, "a" ≠ 0` is ______. (c is arbitrary constant)

Options

• `x/"a" = tan  y/"a" + "c"`

• tan xy = c

• tan (x + y) = c

• `tan  (y + "c")/"a" = (x + y)/"a"`

Answer 1

General solution of `(x + y)^2 ("d"y)/("d"x) = "a"^2, "a" ≠ 0` is `tan  (y + "c")/"a" = (x + y)/"a"`. (c is arbitrary constant)

Explanation:

`("d"y)/("d"x) = "a"^2/(x + y)^2`  ......(i)

Put x + y = v   ......(ii)

⇒ `("d"y)/("d"x) = "dv"/("d"x) - 1`  ......(iii)

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

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

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

⇒ `"v"^2/("a"^2 + "v"^2) "dv" = "d"x`

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

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

Integrating on both sides, we get

`int "dv" - "a"^2 int 1/("a"^2 + "v"^2) "dv" = int "d"x + "c"_1`

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

⇒ `x + y - "a" tan^-1 ((x + y)/"a") = x + "c"_1`

⇒ `(y + "c")/"a" = tan^-1 ((x + y)/"a")`, where c = – c₁

⇒ `tan((y + "c")/"a") = (x + y)/"a"`