The differential equation `y=px+f(p)`, …………..(i)
where `p=(dy)/(dx)`,is known as Clairout's equation. To solve equation i) differentiate it with respect to x, which gives either
`(dp)/(dx)=0 rArr p =c`………….(ii)
or `x+f^(i)(p)=0`…………(iii)
The singular solution of the differential equation `y=mx + m-m^(3)`, where `m=(dy)/(dx)`, passes through the point.

Options

A. (0,0)
B. (0,1)
C. (1,0)
D. (-1,0)

Answer

D

Explanation

`y=mx+m-m^(3)`
Differentiating w.r.t. x, we get
`(dy)/(dx) = m+x (dm)/(dx) + (dm)/(dxm) -3m^(2)(dm)(dx)`
`m=m+x(dm)/(dx) + (dm)/(dx) - 3m^(2)(dm)/(dx)`
`(dm)/(dx)(x+1-3m^(2))=0`
`(dm)/(dx) =0` or `m=c`…………(2)
`(dm)/(dx)=0` or `m=c`
or `x+1-3m^(2)=0` or `m^(2)=(x+1)/3`...............(3)
Eliminating m between equations (1) and (3), we get
`y=m(x+1-m^(2))`
`((x+1)/(3))^(1//2)(x+1-9x+1)/(3)`
`=((x+1)/(3))^(1//2) 2/3(x+1)=2((x+1)/(3))^(3//2)`
or `y^(2)=4/27(x+1)^(3)`
or `27y^(2)=4(x+1)^(3)`