Differential Equation for Circles with Centers on Line Y Equals X

To find the differential equation representing the family of circles whose centers lie on the straight line $y = x$, we start by writing the general equation of such circles.

Let the center of the circle be $(a, a)$ and its radius be $r$. Then, the equation of the circle is given by: $(x - a)^2 + (y - a)^2 = r^2$ where $a$ and $r$ are arbitrary constants.

Differentiating both sides with respect to $x$: $2(x - a) + 2(y - a)\frac{dy}{dx} = 0$ $(x - a) + (y - a)y' = 0 \quad \ldots (1)$

We can express $(x - a)$ in terms of $(y - a)$: $(x - y) + (y - a) + (y - a)y' = 0$ $(x - y) + (y - a)(1 + y') = 0$ $(y - a) = \frac{y - x}{1 + y'} \quad \ldots (2)$

Now, differentiating equation (1) with respect to $x$: $1 + (y')^2 + (y - a)y'' = 0 \quad \ldots (3)$

Substituting the expression for $(y - a)$ from equation (2) into equation (3): $1 + (y')^2 + \left(\frac{y - x}{1 + y'}\right) y'' = 0$

Multiplying the entire equation by $(1 + y')$: $(y - x)y'' + (1 + y')\left(1 + (y')^2\right) = 0$ $(y - x)y'' + \left(1 + y' + (y')^2 + (y')^3\right) = 0$ $(y - x)y'' + \left(1 + y' + (y')^2\right)y' + 1 = 0$

Comparing this with the given differential equation: $P y'' + Q y' + 1 = 0$

We obtain: $P = y - x$ $Q = 1 + y' + (y')^2$

Now, evaluating the given options:

1. Option (A) and (B): $P = y - x$ Hence, statement (B) is true, and (A) is false.

2. Option (C): $P + Q = (y - x) + \left(1 + y' + (y')^2\right) = 1 - x + y + y' + (y')^2$ Hence, statement (C) is true.

3. Option (D): $P - Q = (y - x) - \left(1 + y' + (y')^2\right) = -1 - x + y - y' - (y')^2 \neq x + y - y' - (y')^2$ Hence, statement (D) is false.

Therefore, the correct statements are (B) and (C).