merge main

quarto/ODEs/Project.toml

@@ -1,11 +1,16 @@
 [deps]
 CalculusWithJulia = "a2e0e22d-7d4c-5312-9169-8b992201a882"
 IJulia = "7073ff75-c697-5162-941a-fcdaad2a7d2a"
+LaTeXStrings = "b964fa9f-0449-5b57-a5c2-d3ea65f4040f"
 ModelingToolkit = "961ee093-0014-501f-94e3-6117800e7a78"
 MonteCarloMeasurements = "0987c9cc-fe09-11e8-30f0-b96dd679fdca"
+Mustache = "ffc61752-8dc7-55ee-8c37-f3e9cdd09e70"
 OrdinaryDiffEq = "1dea7af3-3e70-54e6-95c3-0bf5283fa5ed"
 PlotlyBase = "a03496cd-edff-5a9b-9e67-9cda94a718b5"
 PlotlyKaleido = "f2990250-8cf9-495f-b13a-cce12b45703c"
 Plots = "91a5bcdd-55d7-5caf-9e0b-520d859cae80"
+QuizQuestions = "612c44de-1021-4a21-84fb-7261cf5eb2d4"
 Roots = "f2b01f46-fcfa-551c-844a-d8ac1e96c665"
 SymPy = "24249f21-da20-56a4-8eb1-6a02cf4ae2e6"
+Tables = "bd369af6-aec1-5ad0-b16a-f7cc5008161c"
+TextWrap = "b718987f-49a8-5099-9789-dcd902bef87d"

quarto/ODEs/differential_equations.qmd

@@ -411,7 +411,7 @@ p  = [γ => 0.0,
 prob = ODEProblem(sys, u0, TSPAN, p, jac=true)
 sol = solve(prob,Tsit5())
 
-plot(t -> sol(t)[3], t -> sol(t)[4], TSPAN..., legend=false)
+plot(t -> sol(t)[1], t -> sol(t)[3], TSPAN..., legend=false)
 ```
 
 The toolkit will automatically generate fast functions and can perform transformations (such as is done by `ode_order_lowering`) before passing along to the numeric solves.

quarto/ODEs/odes.qmd

@@ -340,7 +340,7 @@ The first-order initial value equations we have seen can be described generally
 $$
 \begin{align*}
 y'(x) &= F(y,x),\\
-y(x_0) &= x_0.
+y(x_0) &= y_0.
 \end{align*}
 $$
 
@@ -375,7 +375,7 @@ $$
 u'(x) = a u(1-u), \quad a > 0
 $$
 
-Before beginning, we look at the form of the equation. When $u=0$ or $u=1$ the rate of change is $0$, so we expect the function might be bounded within that range. If not, when $u$ gets bigger than $1$, then the slope is negative and when $u$ gets less than $0$, the slope is positive, so there will at least be a drift back to the range $[0,1]$. Let's see exactly what happens. We define a parameter, restricting `a` to be positive:
+Before beginning, we look at the form of the equation. When $u=0$ or $u=1$ the rate of change is $0$, so we expect the function might be bounded within that range. If not, when $u$ gets bigger than $1$, then the slope is negative and though the slope is negative too when $u<0$, but for a realistic problem, it always be $u\ge0$. so we focus $u$ on the range $[0,1]$. Let's see exactly what happens. We define a parameter, restricting `a` to be positive:
 
 
 ```{julia}
@@ -403,7 +403,7 @@ To finish, we call `dsolve` to find a solution (if possible):
 out = dsolve(eqn)
 ```
 
-This answer - to a first-order equation - has one free constant, `C_1`, which can be solved for from an initial condition. We can see that when $a > 0$, as $x$ goes to positive infinity the solution goes to $1$, and when $x$ goes to negative infinity, the solution goes to $0$ and otherwise is trapped in between, as expected.
+This answer - to a first-order equation - has one free constant, `C₁`, which can be solved for from an initial condition. We can see that when $a > 0$, as $x$ goes to positive infinity the solution goes to $1$, and when $x$ goes to negative infinity, the solution goes to $0$ and otherwise is trapped in between, as expected.
 
 
 The limits are confirmed by investigating  the limits of the right-hand:
@@ -420,7 +420,7 @@ We can confirm that the solution is always increasing, hence trapped within $[0,
 diff(rhs(out),x)
 ```
 
-Suppose that $u(0) = 1/2$. Can we solve for $C_1$ symbolically? We can use `solve`, but first we will need to get the symbol for `C_1`:
+Suppose that $u(0) = 1/2$. Can we solve for $C_1$ symbolically? We can use `solve`, but first we will need to get the symbol for `C₁`:
 
 
 ```{julia}