Key Concepts and Formulas

• Differential Equation Formation: To form a differential equation from a family of curves, differentiate the equation with respect to the independent variable (usually $x$) as many times as there are arbitrary constants. Then, eliminate the arbitrary constants using the original equation and the derived equations.
• Implicit Differentiation: When differentiating an equation where $y$ is implicitly a function of $x$, use the chain rule. For example, the derivative of $y^2$ with respect to $x$ is $2y \frac{dy}{dx}$ or $2yy'$.
• Notation: $y' = \frac{dy}{dx}$ represents the first derivative of $y$ with respect to $x$.

Step-by-Step Solution

Step 1: Differentiate the given equation with respect to x.

We are given the equation of the family of circles: $x^2 + y^2 - 2ay = 0$ Since there is one arbitrary constant, 'a', we need to differentiate the equation once with respect to $x$. Differentiating both sides with respect to $x$, we get: $\frac{d}{dx}(x^2 + y^2 - 2ay) = \frac{d}{dx}(0)$ $2x + 2y\frac{dy}{dx} - 2a\frac{dy}{dx} = 0$ Using the notation $y' = \frac{dy}{dx}$, we have: $2x + 2yy' - 2ay' = 0$ Dividing by 2: $x + yy' - ay' = 0 \quad \ldots(2)$

Step 2: Solve for 'a' in terms of x, y, and y'.

Our goal is to eliminate 'a' from equations (1) and (2). From equation (2), we can isolate 'a': $ay' = x + yy'$ $a = \frac{x + yy'}{y'} \quad \ldots(3)$

Step 3: Substitute the value of 'a' from equation (3) into equation (1).

Substitute the expression for 'a' from equation (3) into equation (1): $x^2 + y^2 - 2\left(\frac{x + yy'}{y'}\right)y = 0$ $x^2 + y^2 - \frac{2xy + 2y^2y'}{y'} = 0$

Step 4: Simplify the equation to eliminate the fraction.

Multiply both sides of the equation by $y'$: $(x^2 + y^2)y' - (2xy + 2y^2y') = 0$ $x^2y' + y^2y' - 2xy - 2y^2y' = 0$ $x^2y' - y^2y' - 2xy = 0$ $(x^2 - y^2)y' = 2xy$ This does not match the target equation, so let us go back to step 3 and re-examine.

Step 3 (Revised): Solve for 2a from equation (1) and substitute. From equation (1): $2a = \frac{x^2 + y^2}{y}$ Substituting into equation (2) multiplied by 2: $2x + 2yy' - 2ay' = 0$ $2x + 2yy' - \frac{x^2+y^2}{y}y' = 0$ Multiply by $y$ to clear the fraction: $2xy + 2y^2y' - (x^2 + y^2)y' = 0$ $2xy + 2y^2y' - x^2y' - y^2y' = 0$ $2xy + y^2y' - x^2y' = 0$ $2xy = (x^2 - y^2)y'$ This also does not match the target equation. Let's review the original differentiation.

Step 1 (Revised): Differentiate the given equation with respect to x.

We are given the equation of the family of circles: $x^2 + y^2 - 2ay = 0$ Since there is one arbitrary constant, 'a', we need to differentiate the equation once with respect to $x$. Differentiating both sides with respect to $x$, we get: $\frac{d}{dx}(x^2 + y^2 - 2ay) = \frac{d}{dx}(0)$ $2x + 2y\frac{dy}{dx} - 2a\frac{dy}{dx} = 0$ Using the notation $y' = \frac{dy}{dx}$, we have: $2x + 2yy' - 2ay' = 0$ Dividing by 2: $x + yy' - ay' = 0 \quad \ldots(2)$

Step 2 (Revised): Solve for a from equation (2).

From equation (2), we can isolate 'a': $x + yy' = ay'$ $a = \frac{x + yy'}{y'}$

Step 3 (Revised): Solve for 2a from equation (1).

From equation (1): $2a = \frac{x^2 + y^2}{y}$

Step 4: Equate the two expressions for a.

Equating the two expressions for a, we get: $\frac{x^2 + y^2}{2y} = \frac{x + yy'}{y'}$ Cross-multiplying: $(x^2 + y^2)y' = 2y(x + yy')$ $(x^2 + y^2)y' = 2xy + 2y^2y'$ $(x^2 + y^2)y' - 2y^2y' = 2xy$ $(x^2 - y^2)y' = 2xy$

Still not the target answer. Let's solve for a in equation 2 instead and sub into equation 1. From (2), $ay' = x + yy'$, so $a = \frac{x+yy'}{y'}$. Sub into (1): $x^2 + y^2 - 2y(\frac{x+yy'}{y'}) = 0$ Multiply by $y'$: $(x^2 + y^2)y' - 2y(x+yy') = 0$ $(x^2 + y^2)y' - 2xy - 2y^2y' = 0$ $(x^2+y^2-2y^2)y' = 2xy$ $(x^2 - y^2)y' = 2xy$.

We are still not getting the answer.

Let's instead try solving for $y'$ from equation (2): $ay' = x + yy'$, so $y'(a-y) = x$ which means $y' = \frac{x}{a-y}$. Substituting into equation (1): $x^2 + y^2 - 2ay = 0$. Then $2ay = x^2 + y^2$, so $a = \frac{x^2+y^2}{2y}$. Thus $y' = \frac{x}{\frac{x^2+y^2}{2y} - y} = \frac{x}{\frac{x^2+y^2 - 2y^2}{2y}} = \frac{2xy}{x^2-y^2}$. Thus $(x^2-y^2)y' = 2xy$.

The provided correct answer must be wrong. Let's try starting from option (A) and see if we can reconstruct the original equation. $(x^2 + y^2)y' = 2xy$, so $y' = \frac{2xy}{x^2 + y^2}$. Since $y' = \frac{dy}{dx}$, we have $\frac{dy}{dx} = \frac{2xy}{x^2+y^2}$. $\frac{x^2+y^2}{y} dy = 2x dx$. Integrating both sides, $\int \frac{x^2+y^2}{y} dy = \int 2x dx$. $\int (\frac{x^2}{y} + y) dy = x^2 + C$. This does not seem to lead anywhere productive.

Let us reconsider Step 2. We have $x + yy' - ay' = 0$, so $x + yy' = ay'$. From the original equation, $x^2 + y^2 = 2ay$. Then $a = \frac{x^2 + y^2}{2y}$. Substituting into $x + yy' = ay'$, we have $x + yy' = \frac{x^2+y^2}{2y}y'$. $2xy + 2y^2y' = (x^2+y^2)y'$ $2xy = (x^2+y^2 - 2y^2)y'$ $2xy = (x^2 - y^2)y'$ Thus, $(x^2 - y^2)y' = 2xy$.

I suspect there is an error with the problem statement or the answer key. Let us derive the solution again from scratch. $x^2 + y^2 - 2ay = 0$. Differentiating: $2x + 2yy' - 2ay' = 0$. $x + yy' = ay'$. Then $a = \frac{x+yy'}{y'}$. Substituting into the original: $x^2 + y^2 - 2y(\frac{x+yy'}{y'}) = 0$. Multiplying by $y'$: $(x^2 + y^2)y' - 2xy - 2y^2y' = 0$. $(x^2 - y^2)y' = 2xy$.

Common Mistakes & Tips

• Implicit Differentiation: Remember to apply the chain rule correctly when differentiating terms involving $y$ with respect to $x$.
• Algebraic Manipulation: Carefully perform algebraic manipulations to avoid errors when isolating and substituting variables. Double-check your work.
• Checking the Solution: If possible, check your derived differential equation by substituting a known solution from the original family of curves.

Summary

We started with the equation of a family of circles, $x^2 + y^2 - 2ay = 0$, and aimed to find a differential equation that represents this family. We differentiated the equation once with respect to $x$ to eliminate the arbitrary constant 'a'. After substituting the expression for 'a' back into the original equation and simplifying, we obtained the differential equation $(x^2 - y^2)y' = 2xy$. However, this does not match the provided answer. After careful re-derivation, it became apparent that either the question or the provided answer key has an error.

Final Answer

The derived differential equation is $(x^2 - y^2)y' = 2xy$, which does not correspond to any of the provided options. I suspect that there is an error in the problem or in the given correct answer.