Question #74113
2 Answers
Explanation:
This equation represents
We know that
Option (2);
# x^2 + 2xydy/dx = y^2 # is the correct DE
Explanation:
Rather than attempting to solve each DE, let us form a differential equation that fits the given model, and see which solution the DE matches
The general equation of a circle of centre
# (x-a)^2 + (y-b)^2 = r^2 #
For the circle to have it's centre on the
# (x-a)^2 + y^2 = r^2 #
# :. x^2 -2ax+a^2 + y^2 = r^2 \ \ \ \ \ ..... [1] #
For the circle to pass through the origin (ie
# a^2 = r^2 #
and so we can substitute into equation [1] to get
# x^2 -2ax+ y^2 = 0 \ \ \ \ \ ..... [2]#
If we differentiate [2] implicitly we get:
# 2x-2a + 2ydy/dx = 0 \ \ \ \ \ ..... [3]#
And if we rearrange [2] (to eliminate
# 2ax = x^2 + y^2 #
# :. 2a = x + y^2/x #
Substituting this last result into [3] gives us:
# 2x-(x + y^2/x) + 2ydy/dx = 0 ... [3]#
# :. 2x-x - y^2/x + 2ydy/dx = 0 .#
# :. x - y^2/x + 2ydy/dx = 0 #
# :. x^2 - y^2 + 2xydy/dx = 0 #
# :. x^2 + 2xydy/dx = y^2 #
Which matches option (2)
Incidentally the form of the solution for the various DE options are:
# {: (x^2=y^2+3xydy/dx,=> y^2=c/x^(2/3)+x^2/4), (y^2=x^2+2xydy/dx,=>y^2=cx-x^2), (y^2=x^2-2xydy/dx,=> y^2=c/(3x)+x^2/3), (x^2=y^2+xydy/dx,=>y^2=c/(2x^2)+x^2/2) :} #