The differential equation of all ellipses centred at the origin is

Options

A. `y_(2)+xy_(1)^(2)-yy_(1)=0`
B. `xyy_(2)+xy_(1)^(2)-yy_(1)=0`
C. `yy_(2)+xy_(1)^(2)-xy_(1)=0`
D. none of these

Answer

To find the differential equation of all ellipses centered at the origin, we start with the standard equation of an ellipse. The equation of an ellipse centered at the origin with semi-major axis \( a \) and semi-minor axis \( b \) is given by:

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

### Step 1: Differentiate the Equation
We differentiate the equation with respect to \( x \). Using implicit differentiation, we get:

\[ \frac{d}{dx}\left(\frac{x^2}{a^2}\right) + \frac{d}{dx}\left(\frac{y^2}{b^2}\right) = 0 \]

This results in:

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

Let \( y' = \frac{dy}{dx} \). Thus, we can rewrite the equation as:

\[ \frac{2x}{a^2} + \frac{2y y'}{b^2} = 0 \]

### Step 2: Solve for \( y' \)
Rearranging gives:

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

Dividing both sides by 2:

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

### Step 3: Differentiate Again
Next, we differentiate the equation again with respect to \( x \):

\[ \frac{d}{dx}\left(\frac{y y'}{b^2}\right) = \frac{d}{dx}\left(-\frac{x}{a^2}\right) \]

Using the product rule on the left side:

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

### Step 4: Substitute Back
Now, we can substitute \( b^2 \) in terms of \( a^2 \) and \( y \) using the original ellipse equation. From the first derivative, we can express \( b^2 \) as:

\[ b^2 = \frac{y y'}{-\frac{x}{a^2}} \Rightarrow b^2 = -\frac{a^2 y y'}{x} \]

### Step 5: Form the Differential Equation
Substituting \( b^2 \) back into our differentiated equation gives:

\[ \frac{y y''}{-\frac{a^2 y y'}{x}} + \frac{(y')^2}{-\frac{a^2 y y'}{x}} = -\frac{1}{a^2} \]

After simplification, we arrive at the final form of the differential equation:

\[ x y'' + (y')^2 - y y' = 0 \]

### Final Result
Thus, the differential equation of all ellipses centered at the origin is:

\[ x y'' + (y')^2 - y y' = 0 \]

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