Lecture 1 - Introduction

Differential equations and initial conditions

A first-order ordinary differential equation specifies a rate of change:

$$y^{\prime }(t)=f(t,y(t)).$$

$t$ is the independent variable; $y$ is the dependent variable. A solution is a differentiable function satisfying the equation on an interval.

An initial value problem (IVP) adds a starting value:

$$y^{\prime }=f(t,y),y(t_0)=y_0.$$

If the velocity depends only on time, $y^{\prime }=v(t)$, direct integration gives

$$\boxed{y(t)=y_0+{\int }_{t_0}^{t}v(s)ds.}$$

$s$ is a dummy variable; $t$ is the endpoint. For constant velocity, $y(t)=y_0+v(t-t_0)$.

For continuous $f$, the general IVP can be written as

$$y(t)=y_0+{\int }_{t_0}^{t}f(s,y(s))ds.$$

This is an implicit relation: the unknown function still appears inside the integral.

Exponential growth and decay

For constant $a$,

$$y^{\prime }=ay⟹y=C{e}^{at}.$$

Differentiating verifies $(C{e}^{at}{)}^{\prime }=aC{e}^{at}$. The initial condition determines $C$:

$$\boxed{y(t)=y_0{e}^{a(t-t_0)}.}$$

For example, $y(0)=3$ gives $y=3e^{at}$. For nonzero initial data, the magnitude grows when $a>0$ and decays when $a<0$.

Model	Equation	Solution
Radioactive decay, $a>0$	$m^{\prime }=-am$	$m=m_0{e}^{-a(t-t_0)}$
Cooling toward constant room temperature $T_r$, $k>0$	$T^{\prime }=-k(T-T_r)$	$T=T_r+(T_0-T_r)e^{-k(t-t_0)}$
Population growth with unlimited resources, $b>0$	$n^{\prime }=bn$	$n=n_0{e}^{b(t-t_0)}$

Order and time dependence

The order is the highest derivative present, regardless of its power:

$$\begin{array}{rl}{y}^{\prime \prime }=1& \text{second order},\\ \sin (y^{\prime })+\cos (y^{\prime \prime \prime })=0& \text{third order},\\ (y^{\prime \prime }{)}^5+(y^{\prime }{)}^2+\ln y=0& \text{second order}.\end{array}$$

An ODE involves derivatives with respect to one independent variable. A partial differential equation involves partial derivatives, such as $u_t$ and $u_{xx}$ for $u(x,t)$.

An autonomous equation has no explicit time dependence:

$$y^{\prime }=f(y),\text{e.g. }{y}^{\prime }=y^2-\sin (e^y).$$

A non-autonomous equation depends explicitly on time, such as $y^{\prime }=y^2-t$.

Direction fields and equilibria

The direction field assigns a slope $f(t,y)$ to each point $(t,y)$. Solution curves are tangent to these slopes. For an autonomous equation, the slope is constant along each horizontal line.

For $y^{\prime }=\sin y$:

$$\begin{array}{rl}\sin y>0& \text{on }(2k\pi ,(2k+1)\pi ),\\ \sin y<0& \text{on }((2k-1)\pi ,2k\pi ),k\in \mathbb{Z}.\end{array}$$

An equilibrium is a constant solution $y(t)=y_{\ast }$, so

$$f(t,y_{\ast })=0\text{for every }t\text{ in the interval}.$$

Equation	Equilibria
$y^{\prime }=\sin y$	$y_{\ast }=k\pi$, $k\in \mathbb{Z}$
$y^{\prime }=t(y+1)$	$y_{\ast }=-1$
$y^{\prime }=ty+1$	None: no constant $y_{\ast }$ makes $t{y}_{\ast }+1=0$ for every $t$
$y^{\prime }=y^2+1$	None over $\mathbb{R}$

For $y^{\prime }=\sin y$, nearby solutions move toward odd multiples of $\pi$ and away from even multiples.

Next: Lecture 2 - Linear First-Order Differential Equations