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Question
The differential equation of the family of curves x2 + y2 – 2ay = 0, where a is arbitrary constant, is ______.
Options
`(x^2 - y^2) ("d"y)/("d"x)` = 2xy
`2(x^2 + y^2) ("d"y)/("d"x)` = xy
`2(x^2 - y^2) ("d"y)/("d"x)` = xy
`(x^2 + y^2) ("d"y)/("d"x)` = 2xy
Solution
The differential equation of the family of curves x2 + y2 – 2ay = 0, where a is arbitrary constant, is `(x^2 - y^2) ("d"y)/("d"x)` = 2xy.
Explanation:
The given equation is x2 + y2 – 2ay = 0 ......(1)
Differentiating w.r.t. x, we have
`2x + 2y * ("d"y)/("d"x) - 2"a" ("d"y)/("d"x)` = 0
⇒ `x + y ("d"y)/("d"x) - "a" ("d"y)/("d"x)` = 0
⇒ `x + (y - "a") ("d"y)/("d"x)` = 0
⇒ `(y - "a") ("d"y)/("d"x)` = – x
⇒ y – a = `(-x)/(("d"y)/("d"x))`
⇒ a = `y + x/(("d"y)/("d"x))`
⇒ a = `(y * ("d"y)/("d"x) + x)/(("d"y)/("d"x))`
Putting the value of a in equation (1) we get
`x^2 + y^2 - 2y [(y ("d"y)/("d"x) + x)/(("d"y)/("d"x))]` = 0
⇒ `(x^2 + y^2) ("d"y)/("d"x) - 2y(y ("d"y)/("d"x) + x)` = 0
⇒ `(x^2 + y^2) ("d"y)/("d"x) - 2y^2 ("d"y)/("d"x) - 2xy` = 0
⇒ `(x^2 + y^2 - 2y^2) ("d"y)/("d"x^2)` = 2xy
⇒ `(x^2 - y^2) ("d"y)/("d"x)` = 2xy
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