The differential equation of all circles passing through origin and having their centres on the x-axis is ........

Answer

To find the differential equation of all circles passing through the origin and having their centers on the x-axis, we can follow these steps:

### Step 1: Write the equation of the circle
The general equation of a circle with center at point \( (h, 0) \) (since the center lies on the x-axis) and radius \( r \) is given by:
\[ (x - h)^2 + (y - 0)^2 = r^2 \]
Since the circle passes through the origin \( (0, 0) \), we can substitute these coordinates into the equation:
\[ (0 - h)^2 + (0 - 0)^2 = r^2 \]
This simplifies to:
\[ h^2 = r^2 \]
Thus, we can express the radius \( r \) in terms of \( h \):
\[ r = h \]

### Step 2: Substitute the radius into the circle's equation
Now substituting \( r = h \) back into the equation of the circle, we have:
\[ (x - h)^2 + y^2 = h^2 \]

### Step 3: Differentiate the equation
Next, we differentiate the equation with respect to \( x \):
\[ \frac{d}{dx}[(x - h)^2 + y^2] = \frac{d}{dx}[h^2] \]
Using the chain rule, we get:
\[ 2(x - h)(1 - \frac{dh}{dx}) + 2y\frac{dy}{dx} = 0 \]
Dividing through by 2 simplifies this to:
\[ (x - h)(1 - \frac{dh}{dx}) + y\frac{dy}{dx} = 0 \]

### Step 4: Solve for \( h \)
Rearranging the above equation gives:
\[ (x - h)(1 - \frac{dh}{dx}) = -y\frac{dy}{dx} \]
Now, we can isolate \( h \):
\[ x - h = -\frac{y\frac{dy}{dx}}{1 - \frac{dh}{dx}} \]
Thus,
\[ h = x + \frac{y\frac{dy}{dx}}{1 - \frac{dh}{dx}} \]

### Step 5: Substitute \( h \) back into the circle equation
We substitute this expression for \( h \) back into the original circle equation:
\[ (x - (x + y\frac{dy}{dx}))^2 + y^2 = (x + y\frac{dy}{dx})^2 \]
This simplifies to:
\[ (-y\frac{dy}{dx})^2 + y^2 = (x + y\frac{dy}{dx})^2 \]

### Step 6: Expand and simplify
Expanding both sides:
\[ y^2\left(\frac{dy}{dx}\right)^2 + y^2 = x^2 + 2xy\frac{dy}{dx} + y^2\left(\frac{dy}{dx}\right)^2 \]
The \( y^2\left(\frac{dy}{dx}\right)^2 \) terms cancel out:
\[ y^2 = x^2 + 2xy\frac{dy}{dx} \]

### Step 7: Rearranging to form the differential equation
Rearranging gives us:
\[ x^2 - y^2 + 2xy\frac{dy}{dx} = 0 \]

### Final Result
Thus, the differential equation of all circles passing through the origin and having their centers on the x-axis is:
\[ x^2 - y^2 + 2xy\frac{dy}{dx} = 0 \]

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