Find the differential equation representing the family of curves $\mathrm{y}={\mathrm{ae}}^{\mathrm{bx}+5}$, where a and b are arbitrary constants.

Given: curve $\mathrm{y}={\mathrm{ae}}^{\mathrm{bx}+5}$.....(1)
Differentiate that w.r.t. x, it follows
$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{ae}}^{\mathrm{bx}+5}\cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{bx}+5)\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{ae}}^{\mathrm{bx}+5\cdot \mathrm{b}} ....\left(\mathrm{ii}\right)\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}\cdot \mathrm{b} [\mathrm{From} (\mathrm{i}\left)\right]\end{array}$
Differentiating again w.r.t. x, we get
$\begin{array}{l}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\frac{\mathrm{dy}}{\mathrm{dx}}\cdot \mathrm{b}\\ \Rightarrow \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\frac{\mathrm{dy}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\cdot \frac{1}{\mathrm{y}}\right)\left[\because \mathrm{b}=\frac{1}{\mathrm{y}}\frac{\mathrm{dy}}{\mathrm{dx}}\right]\\ \Rightarrow \mathrm{y}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\end{array}$
Therefore, the differential equation is $\mathrm{y}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2$.