Question

Find the differential equation of the following:

y = c(x – c)²

Answer 1

y = c(x – c)²  .......(1)

y = c(x² – 2cx + c²)

y = cx² – 2c²x + c³

Differentiating w.r. to x

`("d"y)/("d"x) = "c"[2x] - 2"c"^2(1) + 0`

`("d"y)/("d"x) = 2"c"x - 2"c"^2`

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

From (1) and (2)

`y/((("d"y)/("d"x))) = ("c"(x - "c")^2)/(2"c"(x - "c"))`

`y/((("d"y)/("d"x))) = ((x - "c"))/2`

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

c = `x - [(2y)/((("d"y)/("d"x)))]`

Substituting this value of c and (x – c) in (1), we get

y = `x - [(2y)/((("d"y)/("d"x)))] [(2y)/((("d"y)/("d"x)))]^2`

y = `((x ("d"y)/("d"x) - 2y))/((("d"y)/("d"x))) [(4y^2)/(("d"y)/("d"x))^2]`

y = `((x("d"y)/("d"x)- 2y)(4y^2))/(("d"y)/("d"x))^3`

⇒ `(("d"y)/("d"x))^3 = ((x ("d"y)/("d"x) - 2y)(4y^2))/y`

⇒  `(("d"y)/("d"x))^3 = (x ("d"y)/("d"x) - 2y) (4y^2)`

`(("d"y)/("d"x))^3 = 4xy ("d"y)/("d"x) + 8y^2`

or

`(("d"y)/("d"x))^3 - 4xy ("d"y)/("d"x) + 8y^2` = 0