From the differential equation of the family of circles in the second quadrant and touching the coordinate axes.

Answer

To derive the differential equation of the family of circles in the second quadrant that touch the coordinate axes, we can follow these steps:

### Step 1: Understand the Circle's Position
The circles in the second quadrant that touch the coordinate axes have their centers located at \((-a, a)\), where \(a\) is a positive constant. The radius of each circle is equal to \(a\).

### Step 2: Write the Equation of the Circle
The general equation of a circle with center \((-a, a)\) and radius \(a\) is given by:
\[ (x + a)^2 + (y - a)^2 = a^2 \]

### Step 3: Expand the Circle's Equation
Expanding the equation:
\[ (x + a)^2 + (y - a)^2 = a^2 \]
\[ x^2 + 2ax + a^2 + y^2 - 2ay + a^2 = a^2 \]
\[ x^2 + y^2 + 2ax - 2ay + a^2 = 0 \]

### Step 4: Differentiate the Equation
Now, we differentiate the equation with respect to \(x\):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) + \frac{d}{dx}(2ax) - \frac{d}{dx}(2ay) + \frac{d}{dx}(a^2) = 0 \]
This gives us:
\[ 2x + 2y \frac{dy}{dx} + 2a - 2a \frac{dy}{dx} = 0 \]
Rearranging terms:
\[ 2x + (2y - 2a) \frac{dy}{dx} = 0 \]

### Step 5: Solve for \(a\)
From the differentiated equation, we can isolate \(a\):
\[ (2y - 2a) \frac{dy}{dx} = -2x \]
\[ a = y + \frac{x}{\frac{dy}{dx}} \]

### Step 6: Substitute \(a\) Back into the Circle's Equation
Now, substitute \(a\) back into the original equation:
\[ (x + (y + \frac{x}{\frac{dy}{dx}}))^2 + (y - (y + \frac{x}{\frac{dy}{dx}}))^2 = (y + \frac{x}{\frac{dy}{dx}})^2 \]

### Step 7: Simplify the Equation
After substituting and simplifying, we arrive at the final differential equation:
\[ xy' + yy' + (x + y)^2 = (xy')^2 \]

### Final Answer
The differential equation of the family of circles in the second quadrant that touch the coordinate axes is:
\[ xy' + yy' + (x + y)^2 = (xy')^2 \]