Find the differential equation for the family of curves given by $\left(y - b\right)^2 = 4 \left(x - a\right)$ where a and b are arbitrary constants.
The given equation is $\left(y - b\right)^2 = 4 \left(x - a\right)$
i.e. $y^2 - 2 b y + b^2 = 4x - 4a$ $\;\;\; \cdots \; (1)$
Differentiating equation $(1)$ w.r.t x gives
$2 \; y \; \dfrac{dy}{dx} - 2 \; b \; \dfrac{dy}{dx} = 4$
i.e. $y \; \dfrac{dy}{dx} - b \; \dfrac{dy}{dx} = 2$ $\;\;\; \cdots \; (2)$
Differentiating equation $(2)$ w.r.t x gives
$y \; \dfrac{d^2 y}{dx^2} + \left(\dfrac{dy}{dx}\right)^2 - b \; \dfrac{d^2 y}{dx^2} = 0$
i.e. $\left(y - b\right) \; \dfrac{d^2 y}{dx^2} + \left(\dfrac{dy}{dx}\right)^2 = 0$ $\;\;\; \cdots \; (3)$
Now, from equation $(2)$ we have,
$\left(y - b\right) \; \dfrac{dy}{dx} = 2$
i.e. $\left(y - b\right) = \dfrac{2}{dy / dx}$ $\;\;\; \cdots \; (4)$
Substituting equation $(4)$ in equation $(3)$ gives
$\dfrac{2}{dy / dx} \; \left(\dfrac{d^2 y}{dx^2}\right) + \left(\dfrac{dy}{dx}\right)^2 = 0$
i.e. $2 \; \dfrac{d^2 y}{dx^2} + \left(\dfrac{dy}{dx}\right)^3 = 0$
is the required differential equation.