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प्रश्न
The differential equation of the family of curves y2 = 4a(x + a) is ______.
विकल्प
`y^2 - 4 ("d"y)/("d"x)(x + ("d"y)/("d"x))`
`2y ("d"y)/("d"x)` = 4a
`y ("d"^2y)/("d"x^2) + (("d"y)/("d"x))^2` = 0
`2x ("d"y)/("d"x) + y(("d"y)/("d"x))^2 - y`
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उत्तर
The differential equation of the family of curves y2 = 4a(x + a) is `2x ("d"y)/("d"x) + y(("d"y)/("d"x))^2 - y`.
Explanation:
The given equation of family of curves is y2 = 4a(x + a)
⇒ y2 = 4ax + 4a2 .......(1)
Differentiating both sides, w.r.t. x, we get
`2y * ("d"y)/("d"x)` = 4a
⇒ `y * ("d"y)/("d"x)` = 2a
⇒ `y/2 ("d"y)/("d"x)` = a
Now, putting the value of a in equation (1) we get
`y^2 = 4x(y/2 ("d"y)/("d"x)) + 4(y/2 * ("d"y)/("d"x))^2`
⇒ `y^2 = 2xy ("d"y)/("d"x) + y^2 (("d"y)/("d"x))^2`
⇒ y = `2x ("d"y)/("d"x) + y(("d"y)/("d"x))^2`
⇒ `2x * ("d"y)/("d"x) + y * (("d"y)/("d"x))^2 - y` = 0
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