Find the equation of the curve passing through origin if the slope of the tangent to the curve at any point `(x ,y)i s` equal to the square of the difference of the abscissa and ordinate of the point.

Explanation

Slope of tangent to the curve = `(dy)/(dx)`
and difference of abscissa and ordinate = x-y
According to the question, `" "(dy)/(dx)=(x-y)^(2)" "`...(i)
Put `" "x-y=z`
`rArr" "1-(dy)/(dx)=(dz)/(dx)`
`rArr" "(dy)/(dx)=1-(dz)/(dx)`
On subsituting these values in Eq. (i), we get
`" "1-(dz)/(dx)=z^(2)`
`rArr" "1-z^(2)=(dz)/(dx)`
`rArr" "dx=(dz)/(1-z^(2))` ltBrgt On integrating both sides, we get
`" "intdx=int(dz)/(1-z^(2))`
`rArr" "x=(1)/(2)log|(1+z)/(1-z)|+C`
`rArr" "tx=(1)/(2)log|(1+x-y)/(1-x+y)|+C" "`...(ii)
Since, the curve passes through the origin.
`therefore" "0=(1)/(2)log|(1+0-0)/(1-0+0)|+C`
`rArr" "C=0`
On substituting the value of C in Eq. (ii), we get
`" "x=(1)/(2)log|(1+x-y)/(1-x+y)|`
`rArr" "2x=log|(1+x-y)/(1-x+y)|`
`rArr" "e^(2x)=|(1+x-y)/(1-x+y)|`
`rArr" "(1-x+y)e^(2x)=1+x-y`