Question #74113

2 Answers
Feb 22, 2017

#(2)#

Explanation:

#y^2= x^2+2 x y (dy)/(dx)#. Solving it gives

#y = pm i sqrt(x^2-x C)#. Squaring both sides

#y^2= -x^2+xC# or

#y^2+x^2=C x#

This equation represents

#(x-x_0)^2+y^2= r^2->x^2+y^2-2x_0 x = r^2-x_0^2#

We know that #x_0^2= r^2# and #C = 2x_0# so is is the option (2)

Feb 22, 2017

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 #(a,b)# and radius #r# is

# (x-a)^2 + (y-b)^2 = r^2 #

For the circle to have it's centre on the #x#-axis then #b=0# and so the equation reduces to;

# (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 #x=y=0#) we also have;

# 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 #a#) we get:

# 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) :} #