alexandria/documents/by-course/math-4b/course-notes/main.typ

#import "./dvd.typ": *

#show: dvdtyp.with(
  title: "Math 4B Course Notes",
  author: "Youwen Wu",
  date: "Winter 2025",
  subtitle: [Taught by Guofang Wei],
)

#outline()

= Course logistics

The textbook is Elementary Differential Equations, 11th edition, 2017. Chapters
1-4, 6, 7, and 9 will be covered.

Attendance to discussion sections is mandatory.

= Lecture #datetime(day: 7, month: 1, year: 2025).display()

== Trivial preliminaries

#definition[
  An ODE involves an unknown function of a single variable and its derivatives
  up to some fixed order. The order of an ODE is the order of the highest
  derivative that appears.
]

#example[First order ODE][
  $ (dif y) / (dif x) = y^2 $
]

#example[Second order ODE][
  $
    y'' &= x \
    integral.double y'' dif x &= integral.double x dif x \
    y &= 1 / 6 x^3 + C
  $
]

#definition[
  A function $y(x)$ defined on $(a,b)$ is a *solution* of the ODE
  $ y' = F(x,y) $
  if and only if
  $ y'(x) = F(x, y(x)), forall x in (a,b) $
]

#problem[
  Check that $y(x) = 20 + 10e^(-x / 2)$ is a solution to the ODE
  $ y' = -1 / 2 y + 10 $
]

#definition[
  A first order ODE
  $ y' = F(x,y) $
  is called *linear* if there are functions $A(x)$ and $B(x)$ such that
  $ F(x,y) = A(x) y + B(x) $
]

#example("Linear ODEs")[
  - $y' = x$
  - $y' = y$
  - $y' = x^2$
]

#example("Nonlinear ODEs")[
  - $y' = y^2$
]

#definition[
  In general, a differential equation is called linear if and only if it can be
  written in the form
  $
    a_n (t) (dif^n y) / (dif t^n) + a_(n-1) (t) (d^(n-1) y) / (d t^(n-1)) + dots + a_1 (t) (dif y) / (dif t) + a_0 (t) y = g(t)
  $
  where $a_k (t)$ and $g(t)$ are single variable functions of $t$.
]

#definition[
  *Equilibrium solutions* for the ODE
  $ y' = F(x,y) $
  are solutions $y(x)$ such that $y'(x) = 0$, that is, $y(x)$ is constant.
]

#example[
  The equation
  $ y' = y(y +2) $
  has two equilibria
  $
    y(x) &= 0 \
    y(x) &= -2
  $
]

#problem[
  What are the equilibria of the equation
  $ y' = y(y - x) $
]

== General solution of a first order linear ODE

We start with the differential equation in standard form

$ (dif y) / (dif t) + p(t) y = g(t) $

where $p(t)$ and $g(t)$ are continuous single variable functions of $t$.

Then let us assume the existence of an *integrating factor* $mu(t)$, such that

$ mu(t) p(t) = mu'(t) $

and then multiplying each term by $mu(t)$ to obtain

$ mu(t) (dif y) / (dif t) + mu(t) p(t) y = mu(t) g(t) $

Then

$ mu(t) (dif y) / (dif t) + mu'(t) y = mu(t) g(t) $

Then recognize that the left side of the equation is the product rule to obtain

$
  (mu(t) y(t))' &= mu(t) g(t) \
  integral (mu(t) y(t))' dif t &= integral mu(t) g(t) dif t \
  mu(t) y(t) + C &= integral mu(t) g(t) dif t \
  y(t) &= (integral mu(t) g(t) dif t + C) / (mu(t))
$

Now we have a general solution but we need to determine $mu(t)$.

$
  mu(t) p(t) &= mu'(t) \
  (mu'(t)) / (mu(t)) &= p(t) \
  (ln mu(t))' &= p(t)
$

So now

$
  integral mu(t) + k &= integral p(t) dif t \
  ln mu(t) &= integral p(t) dif t + k \
  mu(t) = e^(integral p(t) dif t + k) &= k e^(integral p(t) dif t)
$

Now substitute

$
  y(t) &= (integral k e^(integral p(t) dif t) g(t) dif t + C) / (k e^(integral p(t))) \
  &= (k integral e^(integral p(t) dif t) g(t) dif t + C) / (k e^(integral p(t)))
$

Now do some cursed constant manipulation to obtain a final solution with only one arbitrary constant

$
  y(t) = (integral e^(integral p(t) dif t) g(t) dif t + C) / (e^(integral p(t)))
$

#remark[
  The most useful result to us is
  $
    y(t) &= (integral mu(t) g(t) dif t + C) / (mu(t)) \
    mu(t) &= e^(integral p(t) dif t)
  $
  We can easily obtain a solution form for any first order linear ODE simply by
  identifying $p(t)$ and $g(t)$.
]