Form the differential equation of the family of curves y^2-2xy+x^2=a^2, a=arbitary const.
I really need to know the answer!!
this is just (x-y)^2 = a^2
the family of curves is just two parallel lines.
that help?
@steve... uhm.. no... its supposed to be a differential equation.... is it ok if we only do one step differential n eliminate a ?
I think this discussion may help. Scroll down some.
http://math.stackexchange.com/questions/163720/find-out-the-differential-equation-of-the-following-families-of-curves
To find the differential equation of the family of curves y^2 - 2xy + x^2 = a^2, where a is an arbitrary constant, we need to differentiate the given equation implicitly with respect to x.
First, let's differentiate both sides of the equation with respect to x:
d/dx (y^2 - 2xy + x^2) = d/dx (a^2)
Using the chain rule, we differentiate each term with respect to x:
2yy' - (2xy' + 2yx) + 2x = 0
Simplifying further, we have:
2yy' - 2xy' - 2yx + 2x = 0
Now, we can rearrange the terms to get the differential equation in terms of y and y':
2yy' - 2xy' - 2yx + 2x = 0
2yy' - 2xy' = 2yx - 2x
2(y - x)y' = 2(y - x)
Finally, we can divide both sides of the equation by 2(y - x) to isolate y':
y' = (y - x)/(y - x)
y' = 1
So, the differential equation of the family of curves y^2 - 2xy + x^2 = a^2 is y' = 1.
Please note that the differential equation y' = 1 represents a straight line with a slope of 1, which is the same for all curves in the given family.