Solution to nonlinear ODE with radicals

[davikrehalt]

I am not too familiar with differential equations but am familiar with basic calculus, I came across this equation trying to describe a particular function:
dy/dx =((sqrt((y-x)^2+y^2)-abs(y))/(y-x))*abs(y)/y

Anyway I tried to separate the variables unsuccessfully and using v(x)=y(x)/x with no success, I really tried and couldn't solve it, so some insight would be appreciated. I tried restricting y to be positive, still without a breakthrough. Sorry if it's confusing, can't figure out the math format.

 

[haruspex]

There is a solution y = cx, if that helps. c satisfies a quartic. One of the roots is 1, but you can discount that, leaving a cubic. All the real roots of that are negative.

 

[davikrehalt]

Are you suggesting that I try to solve the differential equation with respect to c? Could you elaborate a little since I fail to see c satisfying a quartic.

 

[haruspex]

Suppose y > 0 in some domain:
dy/dx = (√((y-x)²+y²)-y)/(y-x)
Put y = cx:
c = (√((c-1)²+c²)-c)/(c-1)
c(c-1) + c = √((c-1)²+c²) = c²
(c-1)²+c² = c⁴
(c-1)² = c²(c²-1)
There's a common factor c-1. That root corresponds to y=x, which makes the original equation indeterminate. Otherwise:
(c-1) = c²(c+1)

OTOH, where y < 0:
dy/dx = -(√((y-x)²+y²)+y)/(y-x)
Put y = cx:
c = -(√((c-1)²+c²)+c)/(c-1)
c(c-1) - c = -√((c-1)²+c²) = c² -2c
etc.

 

[davikrehalt]

Suppose y > 0 in some domain:
dy/dx = (√((y-x)²+y²)-y)/(y-x)
Put y = cx:
c = (√((c-1)²+c²)-c)/(c-1).

Thanks for the reply, but I was looking for a more general solution where y is not necessarily a linear function of x, perhaps where c would also be a function of x. It seems that your solution forces the y to be zero when x is zero. If c is a function of x then,

c+x*dc/dx =(√((c-1)²+c²)-c)/(c-1)
x*dc/dx=(√((c-1)²+c²)-c^2)/(c-1)
Then, I am quite at a loss trying to integrate and solve that for c

 

[haruspex]

Thanks for the reply, but I was looking for a more general solution where y is not necessarily a linear function of x,

Of course. I was merely pointing out one set of solutions. These should shed light on the overall behaviour. Other solutions cannot cross the line y = cx. Finding all the real roots for the two cases (y >/< 0) will carve up the plane into separate domains.