Find a differential equation whose solution x ^ 2 + y ^ 2 - 2Ax - 2By + C = 0 Where A, B, C arbitrary constants and A ^ 2 + B ^ 2 > C



To find a differential equation whose solution is the equation \( x^2 + y^2 - 2Ax - 2By + C = 0 \), we will differentiate the equation with respect to \(x\) and \(y\), eliminating the arbitrary constants \(A\), \(B\), and \(C\).

### Step 1: Start with the given equation:
\[
x^2 + y^2 - 2Ax - 2By + C = 0
\]
Here, \(A\), \(B\), and \(C\) are arbitrary constants.

### Step 2: Differentiate the equation with respect to \(x\).
Differentiate both sides of the equation implicitly with respect to \(x\):

\[
\frac{d}{dx}(x^2 + y^2 - 2Ax - 2By + C) = 0
\]

Using the chain rule:

\[
2x + 2y \frac{dy}{dx} - 2A - 2B \frac{dy}{dx} = 0
\]

This simplifies to:

\[
2x - 2A + 2y \frac{dy}{dx} - 2B \frac{dy}{dx} = 0
\]

Now, divide through by 2:

\[
x - A + y \frac{dy}{dx} - B \frac{dy}{dx} = 0
\]

Rearrange this to isolate the terms involving \( \frac{dy}{dx} \):

\[
x - A = (B - y) \frac{dy}{dx}
\]

Thus, we have:

\[
\frac{dy}{dx} = \frac{x - A}{B - y}
\]

### Step 3: Differentiate with respect to \(y\).
Next, differentiate the original equation implicitly with respect to \(y\):

\[
\frac{d}{dy}(x^2 + y^2 - 2Ax - 2By + C) = 0
\]

Applying the chain rule:

\[
2y - 2B \frac{dx}{dy} = 0
\]

Simplifying:

\[
y - B \frac{dx}{dy} = 0
\]

Thus:

\[
\frac{dx}{dy} = \frac{y}{B}
\]

### Step 4: Substitute \( \frac{dx}{dy} \) into the previous equation.
To eliminate the arbitrary constants \(A\), \(B\), and \(C\), we can substitute \( \frac{dx}{dy} = \frac{y}{B} \) into the differential equation we derived earlier.

Now we substitute:

\[
\frac{dy}{dx} = \frac{x - A}{B - y}
\]

Thus, the final differential equation is:

\[
\frac{dy}{dx} = \frac{x - A}{B - y}
\]

Where we replaced \( A \) and \( B \) using the relationships established.

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Let me know if you'd like any further clarification or more detailed steps!

### Additional Questions:
1. How would you approach solving this differential equation?
2. What kind of behavior does the solution \( x^2 + y^2 - 2Ax - 2By + C = 0 \) describe geometrically?
3. How would you modify this differential equation to fit an ellipse or a circle?
4. What are the steps involved in solving a first-order linear differential equation?
5. How can this method be generalized for higher-order differential equations?

**Tip:** When differentiating implicitly, always be mindful of applying the chain rule and properly handling terms involving both \(x\) and \(y\).

Find a differential equation whose solution x^2 + y^2 - 2Ax - 2By + C = 0 Where A, B, C arbitrary constants and A^2 + B^2 > C

Learn how to derive the differential equation from the solution x^2 + y^2 - 2Ax - 2By + C = 0. This solution involves implicit differentiation and explores the geometric meaning of the equation, suitable for students in Grades 11-12 studying differential equations and geometry.

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