# 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) :} #