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The differential equation representing the family of curves `y^2=2c(x+sqrt(c)),` where `c` is a positive parameter, is of (a) order 1 (b) order 2 (c) degree 3 (d) degree 4

A

(a) order 1

B

(b) order 2

C

(c) degree 3

D

(d) degree 4

Text Solution

Verified by Experts

The correct Answer is:
(a,c)
given, `y^(2) =2c(x+sqrt(c)) …(i)`
On differentiating w.r.t.x, we get
`2ydy/dx=2c rArr c=y dy/dx`
on putting this value of c in Eq. (i), we get
`y^(2) =2ydy/dx(x+sqrt(ydy/dx))`
`rArr y=2dy/dx cdot x+2y^(1//2) (dy/dx)^(3//2)`
`rArr y-2xdy/dx=2sqrt(y)(dy/dx)^(3//2)`
`rArr (y-2xdy/dx)^(2) =4y(dy/dx)^(3)`
Therefore, order of this differential equation is 1 and
degree is 3.
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