Question

Show that the differential equation representing the family of curves y² = `2"a"(x + "a"^(2/3))`, where a is a postive parameter, is `(y^2 - 2xy ("d"y)/("d"x))^3 = 8(y ("d"y)/("d"x))^5`

Answer 1

Given y² = `2"a"x + 2"a"^(5/3)`  ......(1)

Differentiating equation (1) w.r.t ‘x’ we get

`2y ("d"y)/("d"x)` = 2a + 0

`("d"y)/("d"x) = (2"a")/(2y)`

`("d"y)/("d"x) = "a"/y`

⇒ a = `y ("d"y)/("d"x)`

Substituting the values of a in equation (1), we get

y² = `2(y ("d"y)/("d"x)) x + 2(y ("d"y)/("d"x))^(5/3)`

y² = `2x  y ("d"y)/("d"x) + 2(y ("d"y)/("d"x))^(5/3)`

`y^2 - 2xy ("d"y)/("d"x) = 2(y ("d"y)/("d"x))^(5/3)`

Taking cubic root on both sides, we get

`(y^2 - 2xy ("d"y)/("d"x))^3 = 8(y ("d"y)/("d"x))^5`

Hence y² = 2ax + 2a³ is a solution of the differential equation

`(y^2 - 2xy ("d"y)/("d"x))^3 = 8(y ("d"y)/("d"x))^5`