Key Concepts and Formulas

• Equation of an Ellipse: The general equation of an ellipse centered at the origin, with foci on either the x-axis or y-axis, is given by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a$ and $b$ are positive constants.
• Differentiation: We will use implicit differentiation to eliminate the arbitrary constants $a^2$ and $b^2$.
• Elimination of Arbitrary Constants: The goal is to differentiate the equation enough times to create a system of equations that allows us to eliminate the arbitrary constants.

Step-by-Step Solution

Step 1: Write the general equation of the ellipse and use the given point.

The general equation of the ellipse is: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ Since the ellipse passes through the point (0, 3), we can substitute these coordinates into the equation: $\frac{0^2}{a^2} + \frac{3^2}{b^2} = 1$ This simplifies to: $\frac{9}{b^2} = 1$ Therefore, $b^2 = 9$. Substituting this back into the general equation, we get: $\frac{x^2}{a^2} + \frac{y^2}{9} = 1$ This equation now only contains one arbitrary constant $a^2$.

Step 2: Differentiate the equation with respect to x.

Differentiating both sides of the equation $\frac{x^2}{a^2} + \frac{y^2}{9} = 1$ with respect to $x$ gives: $\frac{2x}{a^2} + \frac{2y}{9} \frac{dy}{dx} = 0$ Simplifying, we have: $\frac{x}{a^2} + \frac{y}{9} y' = 0$ We can solve for $\frac{1}{a^2}$: $\frac{1}{a^2} = -\frac{y y'}{9x}$

Step 3: Substitute the expression for $\frac{1}{a^2}$ back into the ellipse equation.

Substituting $\frac{1}{a^2} = -\frac{y y'}{9x}$ into the equation $\frac{x^2}{a^2} + \frac{y^2}{9} = 1$, we get: $x^2 \left(-\frac{y y'}{9x}\right) + \frac{y^2}{9} = 1$ Simplifying, we get: $-\frac{x y y'}{9} + \frac{y^2}{9} = 1$ Multiplying by 9, we have: $-x y y' + y^2 = 9$ Rearranging, we get: $y^2 - x y y' = 9$

Step 4: Differentiate again with respect to x to eliminate the constant 9.

Differentiating the equation $y^2 - x y y' = 9$ with respect to $x$ gives: $2y y' - \left(y y' + x (y')^2 + x y y''\right) = 0$ Simplifying, we get: $2y y' - y y' - x (y')^2 - x y y'' = 0$ $y y' - x (y')^2 - x y y'' = 0$ Rearranging, we have: $x y y'' + x (y')^2 - y y' = 0$ Multiplying by -1, we obtain: $-x y y'' - x (y')^2 + y y' = 0$ $x y y'' + x (y')^2 - y y' = 0$

Step 5: Rearrange the equation

Rearranging the terms, we get: $x y y'' + x (y')^2 - y y' = 0$

Common Mistakes & Tips

• Careful Differentiation: Implicit differentiation can be tricky. Make sure to apply the product rule and chain rule correctly.
• Simplifying After Each Step: Simplifying the equation after each differentiation makes the subsequent steps easier and reduces the chance of errors.
• Recognizing the Form: The final differential equation often needs to be rearranged to match one of the given options.

Summary

We started with the general equation of an ellipse centered at the origin and passing through (0,3). This allowed us to express the equation with only one arbitrary constant. We then differentiated the equation twice to eliminate the arbitrary constant and arrive at the differential equation $x y y'' + x (y')^2 - y y' = 0$. This matches option (A).

Final Answer: The final answer is \boxed{xy y'' + x (y')^2 - y y' = 0}, which corresponds to option (A).