Note 1
ODE
Def.
An equation involves one independent variable and its derivatives is called an ode.
An equation involves two or more independent variable and its partial dervatives, we call it pde.
Def.
The order of ode is the highest derivative that appears in the equation.
Ex. (2nd order) $$ y^{\prime\prime}(t) + 2(y^\prime(t))^2 = 2 $$
Remark
We often form a ode as:
$$
F(t, y(t), y^\prime(t), ..., y^{(n)}(t)) = 0
$$
Def.
Linear: If ode in this form:
$$
\begin{aligned}
F(y(t), y^\prime(t), ..., y^{(n)}(t)) = 0\
a_1(t)y(t) + a_2(t)y^\prime(t) + \cdots = 0
\end{aligned}
$$
(no $t$) is linear, then we say its linear.
Counterexample: (not linear) $$ \begin{aligned} y^\prime y = t\ y^\prime + sin(y) = t \end{aligned} $$
Two Examples
Two important examples, Newton's Law and Animal Growth Speed (Field Mice and Owls).
Newton's Law
A mass object falling down, we have: $$ (F =)\ m{dv\over dt} = mg $$ Consider the fore due to air resistance coefficient $\gamma$, we have: $$ m{dv\over dt} = mg - \gamma v $$
This is a first order and linear ode. For ${dv\over dt}=0$, we have the equilibrium solution.
Field Mice and Owls.
The mouse population $p(t)$, the proportionality factor $r$ is the rate constant or growth rate, $k$ is the predation rate. we have $$ {dp\over dt} = rp - k $$ The equilibrium solution is $p(t) = k/r$.
Solution for these two examples
$$ \begin{aligned} {dp\over dt} &= rp - k\ {dp\over dt (p-k/r)} &= {r}\ {1\over dt}{dp\over (p-k/r)} &= {r}\ {1\over dt}{d\ln(p-k/r)} &= {r}\ {dt\over dt}{d\ln(p-k/r)} &= {r}dt\ \int {d\ln(p-k/r)} &= \int {r}dt\ \ln(p-k/r) &= {t\over r}+C\ p-k/r &= e^{ {t\over r}+C} = e^Ce^{rt}\ p &= {k\over r}+e^Ce^{rt}\ p &= {k\over r}+ce^{rt} \end{aligned} $$ For general solution, we replace $c={p_0 - {k\over r} }$, then $$ p(t) = {k\over r}+(p_0 - {k\over r})e^{rt} $$
For Newton's Law: $$ \begin{aligned} m{dv\over dt} &= mg - \gamma v \ m{dv\over dt} &= \gamma({mg\over \gamma} - v) \ {m\over dt} {dv\over {mg\over \gamma} - v} &= \gamma \ -{m\over dt} \ln({mg\over \gamma} - v) &= \gamma\ \int-{m\over dt} \ln({mg\over \gamma} - v)dt &= \int\gamma dt\ -m \ln({mg\over \gamma} - v) &= \gamma t + C\ \ln({mg\over \gamma} - v) &= {-\gamma t\over m} + C\ {mg\over \gamma} - v &= e^{ {-\gamma t\over m} + C} \ v &= {mg\over \gamma} - e^{ {-\gamma t\over m} + C} \ v &= {mg\over \gamma} - ce^{-\gamma t\over m} \ \end{aligned} $$ In general, we have $$ v(t) = {mg\over \gamma} + (v_0 - {mg\over \gamma})e^{-\gamma t / m} $$
Another Chemical Ex.
Consider a mount of xx water $y$ in pound (box?), we have (per unit time) $$ {dy\over dt} = water_{flow\ in} - water_{flow\ out} $$ This ex. is as same form as the 2nd ex. above.