Question

Question #74113

Answer 1

`(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)

Answer 2

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) :} #