The equation of the curve passing through the origin and satisfying the differential equation ` ((dy)/(dx))^2 = ( x- y)^2` is :

Options

A. `e^(2x)(1-x+y)=1+x-y`
B. `e^(2x)(1+x-y)=1-x+y`
C. `e^(2x)(1-x+y)+(1+x-y)=0`
D. `e^(2x)(1+x+y)=1-x+y`

Answer

A

Explanation

We have, `(dy)/(dx)=(x-y)^(2)`
Let `(x-y)=v`. Then,
`1-(dy)/(dx)=(dv)/(dx)rArr (dy)/(dx)=1-(dv)/(dx)`
`therefore" "(dy)/(dx)=(x-y)^(2)`
`rArr" "1-(dv)/(dx)=v^(2)`
`rArr" "1-v^(2)=(dv)/(dx)`
`rArr" "dx=(1)/(1-v^(2))dv`
`rArr" "2intdx=2int(1)/(1-v^(2))dv`
`rArr" "2x=log((1+v)/(1-v))+logC`
`rArr" "C((1+v)/(1-v))=e^(2x)`
`rArr" "C((x-y+1)/(y-x+1))=e^(2x)rArrC(x-y+1)=e^(2x)(y-x+1)`
Taking C = 1, we find that option (a) is correct.