The differential equation of all ellipses with centres at the origin and the ends of one axis of symmetry at `(pm1,0)`, is

Options

A. `(x^(2)-1)y'-xy=0`
B. `(x^(2)+1)y'+xy=0`
C. `xy'+(x^(2)+1)y=0`
D. `(x^(2)-1)y'+(x-1)y'=0`

Explanation

To find the differential equation of all ellipses centered at the origin with the ends of one axis of symmetry at (±1, 0), we start with the standard form of the ellipse equation.

### Step 1: Write the standard equation of the ellipse
The standard equation of an ellipse centered at the origin is given by:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

Given that the ends of one axis of symmetry are at (±1, 0), we have \(a = 1\). Thus, the equation simplifies to:

\[ \frac{x^2}{1^2} + \frac{y^2}{b^2} = 1 \]

This can be rewritten as:

\[ x^2 + \frac{y^2}{b^2} = 1 \]

### Step 2: Differentiate the equation
Next, we differentiate the equation implicitly with respect to \(x\):

\[ \frac{d}{dx}\left(x^2 + \frac{y^2}{b^2}\right) = \frac{d}{dx}(1) \]

This gives us:

\[ 2x + \frac{2y}{b^2} \frac{dy}{dx} = 0 \]

### Step 3: Solve for \(\frac{dy}{dx}\)
Rearranging the equation to solve for \(\frac{dy}{dx}\):

\[ \frac{2y}{b^2} \frac{dy}{dx} = -2x \]

\[ \frac{dy}{dx} = -\frac{b^2 x}{y} \]

### Step 4: Substitute for \(b^2\)
From the original ellipse equation \(x^2 + \frac{y^2}{b^2} = 1\), we can express \(b^2\) in terms of \(x\) and \(y\):

\[ b^2 = \frac{y^2}{1 - x^2} \]

Substituting this into our expression for \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = -\frac{\frac{y^2}{1 - x^2} x}{y} \]

This simplifies to:

\[ \frac{dy}{dx} = -\frac{y x}{1 - x^2} \]

### Step 5: Rearranging the equation
We can rearrange this to get the final form of the differential equation:

\[ (1 - x^2) \frac{dy}{dx} + xy = 0 \]

This is the required differential equation of all ellipses centered at the origin with the ends of one axis of symmetry at (±1, 0).

### Final Answer
The differential equation of all ellipses with centers at the origin and the ends of one axis of symmetry at (±1, 0) is:

\[ (1 - x^2) \frac{dy}{dx} + xy = 0 \]