Formation of differential equations Questions in English

(A) Step $1$: The equation of the line is $x \cos \alpha + y \sin \alpha = p$.
Step $2$: Differentiate with respect to $x$: $\cos \alpha + \frac{dy}{dx} \sin \alpha = 0$.
Step $3$: This gives $\frac{dy}{dx} = - \frac{\cos \alpha}{\sin \alpha} = - \cot \alpha$.
Step $4$: Differentiate again with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx} (- \cot \alpha) = 0$ (since $\alpha$ is a constant).
Step $5$: Thus, the differential equation is $\frac{d^2y}{dx^2} = 0$.

(A) Step $1$: Differentiate $(y - a)^2 = 4(x - b)$ with respect to $x$: $2(y - a) \frac{dy}{dx} = 4$, which simplifies to $(y - a) \frac{dy}{dx} = 2$.
Step $2$: Differentiate again with respect to $x$: $\frac{dy}{dx} \cdot \frac{dy}{dx} + (y - a) \frac{d^2y}{dx^2} = 0$.
Step $3$: From Step $1$, $(y - a) = \frac{2}{dy/dx}$. Substitute this into the equation from Step $2$: $(\frac{dy}{dx})^2 + \frac{2}{dy/dx} \cdot \frac{d^2y}{dx^2} = 0$.
Step $4$: Multiply by $\frac{dy}{dx}$ to clear the fraction: $(\frac{dy}{dx})^3 + 2 \frac{d^2y}{dx^2} = 0$.
Step $5$: The highest order derivative is $\frac{d^2y}{dx^2}$, and its power is $1$. Thus, the degree is $1$.

(A) Step $1$: The equation of the line in normal form is $x \cos \alpha + y \sin \alpha = p$.
Step $2$: Differentiate with respect to $x$: $\cos \alpha + \frac{dy}{dx} \sin \alpha = 0$.
Step $3$: This implies $\frac{dy}{dx} = -\frac{\cos \alpha}{\sin \alpha} = -\cot \alpha$.
Step $4$: Differentiate again with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx}(-\cot \alpha) = 0$ (since $\alpha$ is a constant).
Step $5$: Thus, the differential equation is $\frac{d^2y}{dx^2} = 0$.

(B) The general equation of a parabola with its axis along the $y$-axis is $y = ax^2 + b$.
Since there are two arbitrary constants $a$ and $b$, we differentiate twice.
Differentiating with respect to $x$: $\frac{dy}{dx} = 2ax$.
Differentiating again with respect to $x$: $\frac{d^2y}{dx^2} = 2a$.
From the first derivative, $a = \frac{1}{2x} \frac{dy}{dx}$.
Substitute $a$ into the second derivative: $\frac{d^2y}{dx^2} = 2 \left( \frac{1}{2x} \frac{dy}{dx} \right) = \frac{1}{x} \frac{dy}{dx}$.
Rearranging gives $x \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$.