Form Differential Equation From Real-life Situations

Recap on Direct & Inverse Proportion

If ${dx \over dt}$ is directly proportional to $x$, $$ {dx \over dt} = kx $$

If ${dx \over dt}$ is inversely proportional to the square of $x$, $$ {dx \over dt} = {k \over x^2} $$

Modelling increase/decrease

\begin{align} {dN \over dt} & = -kN, \text{ where } k \text{ is a positive constant} \\ \\ \int {1 \over N} \phantom{.} dt & = \int -k \phantom{.} dt \\ \ln |N| & = -kt + C \\ |N| & = e^{-kt + C} \\ N & = \pm e^{-kt+C} \\ N & = \pm e^C . e^{-kt} \\ \\ \text{Let } & A = \pm e^C, \\ N & = A e^{-kt}, \text{ where } A = \pm e^C \text{ and } C \text{ is an arbitrary constant} \end{align}

Modelling increase & decrease

\begin{align} \text{Death rate} & = 0.5 \text{ thousand per day} \\ \\ \text{Birth rate} & = kx \\ \\ \text{Rate of change of insects, }{dx \over dt} & = kx - 0.5 \\ \\ \int {1 \over kx - 0.5} \phantom{.} dx & = \int 1 \phantom{.} dt \\ {1 \over k} \int {k \over kx - 0.5} \phantom{.} dx & = t + C \\ {1 \over k} \ln |kx - 0.5| & = t + C \\ \ln |kx - 0.5| & = kt + kC \\ |kx - 0.5| & = e^{kt+kC} \\ kx - 0.5 & = \pm e^{kt+kC} \\ kx - 0.5 & = \pm e^{kC} . e^{kt} \\ \\ \text{Let } & A = \pm e^{kC}, \\ kx - 0.5 & = A e^{kt} \\ kx & = A e^{kt} + 0.5 \\ kx & = A e^{kt} + {1 \over 2} \\ x & = {A \over k} e^{kt} + {1 \over 2k}, \text{ where } A = \pm e^{kC} \text{ and } C \text{ is arbitrary constant} \end{align}