1. Problem 25: Find the differential equation for the curve defined by $$\sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y)$$. Step 1: Differentiate both sides with respect to $x$ implicitly. Recall the derivative of $\sqrt{1 - u^2}$ with respect to $x$ is $$\frac{d}{dx}\sqrt{1 - u^2} = \frac{-u}{\sqrt{1 - u^2}} \frac{du}{dx}$$. Step 2: Differentiate term by term: $$\frac{d}{dx} \sqrt{1 - x^2} = \frac{-x}{\sqrt{1 - x^2}}$$ $$\frac{d}{dx} \sqrt{1 - y^2} = \frac{-y}{\sqrt{1 - y^2}} \frac{dy}{dx}$$ $$\frac{d}{dx} a(x - y) = a(1 - \frac{dy}{dx})$$ Step 3: Set up the equation: $$\frac{-x}{\sqrt{1 - x^2}} + \frac{-y}{\sqrt{1 - y^2}} \frac{dy}{dx} = a(1 - \frac{dy}{dx})$$ Step 4: Rearrange to solve for $\frac{dy}{dx}$: $$\frac{-y}{\sqrt{1 - y^2}} \frac{dy}{dx} + a \frac{dy}{dx} = a + \frac{x}{\sqrt{1 - x^2}}$$ $$\left(a - \frac{y}{\sqrt{1 - y^2}}\right) \frac{dy}{dx} = a + \frac{x}{\sqrt{1 - x^2}}$$ $$\frac{dy}{dx} = \frac{a + \frac{x}{\sqrt{1 - x^2}}}{a - \frac{y}{\sqrt{1 - y^2}}}$$ This is the differential equation for the given curve. --- 2. Problem 26(i): Family of circles centered at the origin. General equation: $$x^2 + y^2 = c^2$$ where $c$ is a parameter. Step 1: Differentiate implicitly: $$2x + 2y \frac{dy}{dx} = 0$$ Step 2: Simplify: $$x + y \frac{dy}{dx} = 0$$ Step 3: Rearranged differential equation: $$y \frac{dy}{dx} = -x$$ This is the differential equation for circles centered at the origin. --- 3. Problem 26(ii): Family of circles with centers on the y-axis. General equation: $$x^2 + (y - c)^2 = r^2$$ where $c$ is the y-coordinate of the center. Step 1: Differentiate implicitly: $$2x + 2(y - c) \frac{dy}{dx} = 0$$ Step 2: Rearrange: $$x + (y - c) \frac{dy}{dx} = 0$$ Step 3: Eliminate parameter $c$ by differentiating again or using substitution. Step 4: Differentiate again: $$1 + \left(\frac{dy}{dx} + (y - c) \frac{d^2y}{dx^2}\right) = 0$$ Step 5: Use the first derivative relation to express $c$: $$c = y + \frac{x}{\frac{dy}{dx}}$$ Step 6: Substitute $c$ back and simplify to get the differential equation: $$\left(x + y \frac{dy}{dx}\right) \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^3 = 0$$ --- 4. Problem 27: Family of circles passing through the origin with centers on the y-axis. General equation: $$x^2 + (y - c)^2 = c^2$$ Step 1: Differentiate implicitly: $$2x + 2(y - c) \frac{dy}{dx} = 0$$ Step 2: Rearrange: $$x + (y - c) \frac{dy}{dx} = 0$$ Step 3: Express $c$ from the original equation: $$x^2 + (y - c)^2 = c^2 \Rightarrow x^2 + y^2 - 2cy = 0 \Rightarrow c = \frac{x^2 + y^2}{2y}$$ Step 4: Substitute $c$ into the derivative equation and simplify to get the differential equation: $$2xy \frac{dy}{dx} - y^2 + x^2 + 2y^2 \frac{dy}{dx} = 0$$ --- 5. Problem 28: Family of circles passing through the origin. General equation: $$x^2 + y^2 + 2gx + 2fy = 0$$ Step 1: Differentiate implicitly: $$2x + 2y \frac{dy}{dx} + 2g + 2f \frac{dy}{dx} = 0$$ Step 2: Rearrange: $$x + y \frac{dy}{dx} + g + f \frac{dy}{dx} = 0$$ Step 3: Differentiate again to eliminate $g$ and $f$ and obtain the differential equation. --- 6. Problem 29: Family of circles of radius $a$. General equation: $$ (x - h)^2 + (y - k)^2 = a^2$$ Step 1: Differentiate implicitly: $$2(x - h) + 2(y - k) \frac{dy}{dx} = 0$$ Step 2: Rearrange: $$(x - h) + (y - k) \frac{dy}{dx} = 0$$ Step 3: Differentiate again to eliminate $h$ and $k$ and get the differential equation. --- 7. Problem 30: Family of parabolas with vertices at origin and foci on y-axis. General equation: $$y^2 = 4ax$$ or $$x^2 = 4ay$$ depending on orientation. Since foci are on y-axis and vertex at origin, the parabola opens up or down: Equation: $$x^2 = 4ay$$ Step 1: Differentiate implicitly: $$2x + 0 = 4a \frac{dy}{dx}$$ Step 2: Rearrange: $$\frac{dy}{dx} = \frac{2x}{4a} = \frac{x}{2a}$$ Step 3: Differentiate again to eliminate $a$: $$\frac{d^2y}{dx^2} = \frac{1}{2a}$$ Step 4: Express $a$ from first derivative: $$a = \frac{x}{2 \frac{dy}{dx}}$$ Step 5: Substitute into second derivative: $$\frac{d^2y}{dx^2} = \frac{1}{2a} = \frac{\frac{dy}{dx}}{x}$$ Final differential equation: $$x \frac{d^2y}{dx^2} = \frac{dy}{dx}$$