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)`
D. `e^(2x)(1+x+y)=1-x+y`

Answer

A

Explanation

We have, `(dy)/(dx)=pm(x-y)`
`ul("CASE I")" When "(dy)/(dx)=(x-y)`
In this case, we have
`1-(dv)/(dx)=v," where "x-y=v`
`rArr" "(dv)/(dx)=1-v`
`rArr" "(1)/(1-v)dv=dx`
`-log(1-v)=x+logC`
`rArr" "(1-v)^(-1)=Ce^(x) rArr(1)/(1-x+y)=Ce^(x)`
It passes through the origin.
`therefore" "C=1`
Hence, `(1)/(1-x+y)=e^(x)" ...(i)"`
`ul("CASE II")" When "(dy)/(dx)=(-x-y)=y-x.`
In this case, we have
`(du)/(dx)+1=u," where"u=y-x`
`rArr" "(du)/(u-1)=dx`
`rArr" "log(u-1)=x+logC`
`rArr" "u-1=Ce^(x)rArr y-x-1=Ce^(x)`
It passes through the origin.
`therefore" "C=-1`
Hence, `y-x-1=-e^(x)or, x-y+1=e^(x)" ...(ii)"`
From (i) and (ii), we obtain
`(x-y+1)/(1-x+y)=e^(2x)rArr (x-y+1)=(1-x+y)e^(2x)`