The differential equation representing the family of circles touching y-axis at the origin is:

Question

Accepted Answer

To determine why Option 2 is the correct answer, we need to analyze the differential equation representing the family of circles that touch the y-axis at the origin.

### Explanation for Option 2: Non-linear and of first order

1. **Family of Circles Touching the Y-axis at the Origin**:
   - The general equation of a circle with center at $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
   - For the circle to touch the y-axis at the origin, the center must be on the x-axis, i.e., $k = 0$, and the radius must equal the x-coordinate of the center, i.e., $r = h$.
   - Therefore, the equation becomes $(x - h)^2 + y^2 = h^2$.

2. **Differential Equation**:
   - To find the differential equation, we eliminate the parameter $h$.
   - Expanding the circle equation: $x^2 - 2xh + h^2 + y^2 = h^2$.
   - Simplifying gives: $x^2 - 2xh + y^2 = 0$.
   - Differentiating with respect to $x$ gives: $2x - 2h + 2yy' = 0$.
   - Solving for $h$, we get $h = x + yy'$.
   - Substituting back, we eliminate $h$: $x^2 - 2x(x + yy') + y^2 = 0$.
   - Simplifying, we get: $x^2 - 2x^2 - 2xyy' + y^2 = 0$, which simplifies to $x^2 + y^2 = 2xyy'$.

3. **Nature of the Differential Equation**:
   - The equation $x^2 + y^2 = 2xyy'$ is non-linear due to the presence of the product $xyy'$.
   - It is of first order because it involves the first derivative $y'$.

### Explanation for Why Other Options are Incorrect:

1. **Option 1: Linear and of first order**:
   - A linear differential equation of the first order has the form $a(x)y' + b(x)y = c(x)$.
   - The equation $x^2 + y^2 = 2xyy'$ is not linear because of the non-linear term $xyy'$.

2. **Option 3: Linear and of second order**:
   - A second-order differential equation involves second derivatives, which are not present in $x^2 + y^2 = 2xyy'$.
   - Additionally, the equation is not linear.

3. **Option 4: Non-linear and of second order**:
   - The equation $x^2 + y^2 = 2xyy'$ is non-linear, but it is of first order, not second order, as it only involves the first derivative $y'$.

In conclusion, the differential equation representing the family of circles touching the y-axis at the origin is non-linear and of first order, which is why Option 2 is the correct answer.