If the independent variable `x` is changed to `y ,` then the differential equation `x(d^2y)/(dx^2)+((dy)/(dx))^3-(dy)/(dx)=0` is changed to `x(d^2x)/(dy^2)+((dx)/(dy))^2=k` where `k` equals____

Answer

1

Explanation

`(dy)/(dx)=1/(dx//dy),(d^(2)y)/(dx^(2))=d/(dy)(1/(dx//dy)).(dy)/(dx) = -1/(dx//dy)^(3)(d^(2)x)/(dy^(2))`
Hence, `x(d^(2)y)/(dx^(2))+((dy)/(dx))^(3)-(dy)/(dx)=0`
becomes `-x.1/(dx//dy)^(2)(d^(2)x)/(dy^(2))+1/(dx//dy)^(3)-1/(dx//dy)=0`
or `x(d^(2)x)/(dy^(2))-1+((dx)/(dy))^(2)=0` or `x(d^(2)x)/(dy^(2))_+((dx)/(dy))^(2)=1`
`therefore k=1`