Modeling ODES

The 4 steps

• Assumptions

• Mathamatical formulation

• solve(quantitivave+qualitive)

• intepret and compare with data

  • if any of these fail we go back to step 1 with what we learned

Examples

At time t=0 a tank contains $Q_0$ kg of salt dissolved in 100L of water, assume that the water containing $\frac{1}{4}kg$ of salt per litre flows into the tank at a rate of rL/min. the tank is well stirred and has an outlet from which the mixture exits the tank at the same rate r

• find an ivp that describles the total amount of salt Q(t) in kilograms contained within the tank at time t.

$$\begin{array}{r}\text{ we need to find an ode invvolving Q}\\ \frac{dQ}{dt}=\frac{r}{4}-r\frac{Q}{100}\\ r\frac{L}{min}\cdot \frac{1}{4}\frac{kg}{L}=\text{ rate of salt added}\\ r\frac{L}{min}\cdot \frac{Q}{100}\frac{kg}{L}=\text{ rate of salt removed}\\ IVP=\frac{dQ}{dt}=\frac{r}{100}(20-Q),\leftrightarrow Q(0)=Q_0\\ \text{Investigate solutions}\\ \text{ qualitiatively: (equilibrium solution)}\frac{dQ}{dt}=0\\ 0=\frac{r}{100}(25-Q)⟹Q=25\\ 25Kg/100L\\ \frac{\frac{1}{4}kg}{L}\\ \text{Quantitatively}\\ \frac{dQ}{dt}=\frac{r}{100}(25-Q)\\ \text{ we can solve using both integrating factor and seprable equation}\end{array}$$

A rocket of mass $m$ is projected upwards with an intial velocity $v_0$. assuming no air resistance, find the escape velocity

• $F=mg$ will not work as it is only valid at surface of the earth

$$\begin{array}{r}F=k\cdot -\frac{1}{(R+x{)}^2}⟹-\frac{k}{(R+x{)}^2}\\ \text{at surface}\\ x=0\to F=mg\equiv F=-\frac{k}{R^2}⟹-\frac{k}{R^2}=-mg⟹F=-\frac{mg{R}^2}{(R+x{)}^2}\\ F=m\frac{dv}{dt}⟹\frac{dv}{dt}=-\frac{g{R}^2}{(R+x{)}^2}\\ \text{we need to make the equation not depend on 3 varibles, we can just use the chain rule}\\ \frac{dv}{dt}=\frac{dv}{dx}\cdot \frac{dx}{dt}=v\frac{dv}{dx}\text{ this doesnt have t in it :)}\\ v\frac{dv}{dx}=-\frac{g{R}^2}{(R+x{)}^{2,}},v(0)=v_0\\ \int vdv=\int \frac{-g{R}^2}{(R+n{)}^2}dx\\ ⟹\frac{v^2}{2}=\frac{g{R}^2}{x+R}+C\\ ⟹\frac{v^2}{v}=\frac{g{R}^2}{x+R}+\frac{r_0^2}{2}-gR\\ \text{ we solve for v=0 to find the maximum positon that the rocket will go}\\ v=0,⟹\frac{g{R}^2}{x+R}=gR-\frac{v_0^2}{2}⟹v_0=\sqrt{r{t}_{2}gR}\\ \text{ because as x approches inf (because the rocket is leaving)x is 0}\end{array}$$