Find the differential equation of the family of all circles which pass through the origin and whose centre lie on y-axis.

Answer

To find the differential equation of the family of all circles that pass through the origin and whose centers lie on the y-axis, we can follow these steps:

### Step 1: Write the equation of the circle
The center of the circle lies on the y-axis, so we can denote the center as (0, k), where k is the y-coordinate. The radius of the circle is the distance from the center to the origin (0, 0), which is k. Therefore, the equation of the circle can be written as:

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

### Step 2: Expand the equation
Expanding the equation gives:

\[ x^2 + (y^2 - 2yk + k^2) = k^2 \]

This simplifies to:

\[ x^2 + y^2 - 2yk + k^2 - k^2 = 0 \]

Thus, we have:

\[ x^2 + y^2 - 2yk = 0 \]

### Step 3: Rearrange the equation
Rearranging the equation leads to:

\[ x^2 + y^2 = 2yk \]

Now, we can express k in terms of x and y:

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

### Step 4: Differentiate with respect to x
Next, we differentiate both sides with respect to x. Using the quotient rule for differentiation, we have:

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

Applying the quotient rule \( \frac{d}{dx}(v/u) = \frac{u \frac{dv}{dx} - v \frac{du}{dx}}{u^2} \):

Let \( v = x^2 + y^2 \) and \( u = 2y \).

Differentiating \( v \):

\[ \frac{dv}{dx} = 2x + 2y\frac{dy}{dx} \]

Differentiating \( u \):

\[ \frac{du}{dx} = 2\frac{dy}{dx} \]

Now applying the quotient rule:

\[ \frac{dk}{dx} = \frac{(2y)(2x + 2y\frac{dy}{dx}) - (x^2 + y^2)(2\frac{dy}{dx})}{(2y)^2} \]

### Step 5: Set the derivative equal to zero
Since k is a constant (the y-coordinate of the center), we have:

\[ \frac{dk}{dx} = 0 \]

Thus, we set the numerator equal to zero:

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

### Step 6: Simplify the equation
Expanding and simplifying gives:

\[ 4xy + 4y^2\frac{dy}{dx} - 2(x^2 + y^2)\frac{dy}{dx} = 0 \]

Rearranging terms leads to:

\[ 4xy + (4y^2 - 2x^2 - 2y^2)\frac{dy}{dx} = 0 \]

This simplifies to:

\[ 4xy + (2y^2 - 2x^2)\frac{dy}{dx} = 0 \]

### Step 7: Solve for the differential equation
Dividing through by 2 gives:

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

Rearranging leads to the final form of the differential equation:

\[ x^2 - y^2 \frac{dy}{dx} - 2xy = 0 \]

### Final Answer
The required differential equation is:

\[ x^2 - y^2 \frac{dy}{dx} - 2xy = 0 \]