Midterm topics:

1. Separable, exact, linear first order ODEs.
2. Integrating factors (for linear first order ODEs and exact ODEs)
3. Direction fields and solution sketching
4. Existence and Uniqueness Theorem for first-order ODEs
5. General theory of second-order linear equations: homogenous vs. non-homogenous equations, existence-uniqueness theorem, principle of superposition, structure of general solutions
6. Wronskians and fundamental sets of solutions
7. Reduction of order method for finding fundamental sets of solutions.
8. Characteristic polynomials for constant coefficient equations
9. Solving non-homogenous equations using the method of undetermined coefficients and variation of parameters
10. Mechanical vibrations: know basic terminology and the equations of motions, but not formulas for amplitudes and frequency gain

Prerequisites

• Complex calculus

What is a differential equation?

A differential equation describes the relationship between a function and its derivatives. ¹

For example, given $y(x)$, the equation

$$\frac{dy}{dx} = \frac{1}{1 + x^2}$$

is a differential equation.

Descriptors of differential equations

There are two types of differential equations:

1. Ordinary differential equations (ODEs) are of one variable.
2. Partial differential equations (PDEs) are of many variables. The equations contain partial derivatives.

The order of a differential equation is the order of the highest derivative that appears in the equation¹.

The solution is given by the continuous function which satisfies the relationship described by the differential equation. Solutions often exist only in certain intervals.

Physical problems that can be described by differential equations are based on time-series functions. The initial condition is usually $t=0$ but must be $t \ge 0$. The functions are of time and other parameters, and their derivatives.

Because the physical equations governing the mass-spring system and the linear circuit are both linear, the resulting ODEs are also linear!

First-order ODEs

First-order ODEs are in three classes:

1. Separable
2. Exact
3. Linear

Any first-order ODE can be rewritten in the form:

$$\begin{equation}\frac{dy}{dt} = f(t, y)\end{equation}$$

where $f$ is a function of the independent variable $t$ and the function $y(t)$.

For example, the ODE

$$3y^2\sin{(t+y)} - \frac{dy}{dt} = 0$$

becomes

$$\frac{dy}{dt} = 3y^2\sin{(t+y)}$$

with $f(t, y) = 3y^2\sin{(t+y)}$

Direction fields

A direction field for equations of form (1) can be constructed by evaluating $f$ at each point of a rectangular grid. At each point, draw a short line segment whose slope is the value of $f$. Each line segment will be tangent to the graph of the solution passing through that point.

Separable equations

A first-order ODE is separable if $f$ can be written as a product of two functions, each of one variable:

$$\frac{dy}{dt} = f,\quad f(t, y) = g(t)\cdot h(y)$$

Such equations can be solved by integrating both sides:

$$\begin{align} &\frac{dy}{dt} = g(t)\cdot h(y)\\ &\implies \frac{1}{h(y)}\frac{dy}{dt} = g(t)\\ &\implies \int\frac{1}{h(y)} \frac{dy}{\bcancel{dt}}\, \bcancel{dt} = \int g(t)\, dt \\ &\implies \int \frac{1}{h(y)}\, dy = \int g(t)\, dt \end{align}$$

The result of the integration is a function of $y$ on the left and a function of $t$ on the right. Solving for $y$ gives us our solution.

Linear equations

A first-order ODE is linear if it is of the form

$$\frac{dy}{dt} + p(t)y = q(t)$$

To solve, we can multiply the above by the non-zero function $\mu(t)$. $\mu$ must satisfy the separable ODE:

$$\frac{d\mu}{dt} = \mu p$$

The reason why is because then, by the product rule:

$$\frac{d}{dt}(\mu y) = \mu y' + \mu' y$$

Which is equal to the left-hand side of the original equation.

We can then integrate both sides:

$$\begin{align} &\int \frac{d}{dt}(\mu y) \, dt = \int q(t)\, dt \\ &\implies \mu y = \int q(t)\, dt + C\end{align}$$

Exact equations

We can solve all first-order differential equations which can be put in the form

$$\frac{d}{dt} \phi(t, y) = 0$$

We can integrate both sides to get

$$\phi(t, y) = C$$

Equations that can be put in this form but are neither separable nor linear are exact.

How can we determine if an equation can be put in the above form? Note that by the chain rule of partial differentiation

$$\frac{d}{dt}\phi(t, y(t)) = \frac{\partial\phi}{\partial t} + \frac{\partial \phi}{\partial y} \frac{dy}{dt}$$

Given a differential equation in most general form

$$M(t, y) + N(t, y) \frac{dy}{dt} = 0$$

It can only be written in the above form if and only if there exists a function $\phi(t, y)$ such that $M(t, y) = \frac{\partial \phi}{\partial t}$ and $N(t, y) = {\partial\phi\over\partial y}$.

The differential equation

$$M(t, y) + N(t, y)\frac{dy}{dt} = 0$$

is said to be exact if ${\partial M\over\partial y} = {\partial N\over\partial t}$.

The reason for this definition, of course, is that the left-hand side ... is the exact derivative of a known function of $t$ and $y$ ¹

To solve any exact equation, note that if $M$ is the partial derivative of $\phi(t, y)$, then

$$\phi(t, y) = \int M\, dt + g(y)$$

Then we may find $g(y)$ by taking the partial derivative of the above with respect to $y$ and notice that that is equal to $N$:

$$\begin{align} N &= {\partial\phi \over \partial y} \\ &= {\partial\over\partial y}\left [ \int M\, dt\right ] + g'(y)\end{align}$$

By comparing terms we may isolate $g'(y)$ and integration gives us our answer $\phi(t, y)$. Setting $\phi$ to zero, we have our solutions.

$$\phi(t, y) = 0$$

Integrating factors for exact equations

Some first-order ODEs are not exact but can be made exact using an integrating factor.

Given the ODE of form $M + N\frac{dy}{dt} = 0$, $\mu(t,y)$ is an integrating factor if

$$\mu M + \mu N \frac{dy}{dt} = 0$$

is exact, with the restriction that the solution of the ODE does not contain $\mu(t,y) = 0$.

The harmonic oscillator

Homogenous and inhomogenous second order linear ODEs with constant coefficients model the harmonic oscillator.

A harmonic oscillator that loses energy during operation is called a damped oscillator. This might happen due to friction (damping) in mechanical systems or due to resistance in circuits.

Depending on the extent, the system can:

• Oscillate with a frequency lower than the undamped case, and an amplitude decreasing in time. Such a system is called an underdamped oscillator.
• Decay to the equilibrium position without oscillations. Such a system is called an overdamped oscillator

If a harmonic oscillator is undamped, it exhibits simple harmonic motion (SHM). This is sinusoidal oscillation with constant amplitude and frequency.

Free vibrations (SHM)

If a system has no damping (which is proportional to velocity) and no external forces, then it exhibits simple harmonic motion. This is the case where $\alpha = 0$ on the velocity term, such that the differential equation can be reduced to the form:

$$x'' + \omega_0^2 x = 0$$

where $\omega_0^2 = \frac{k}{m}$. For a mechanical system, this motion is independent of mass (which can be cancelled out).

Take the characteristic polynomial $X(r) = r^2 + \omega_0^2$ such that

$$X(r) = 0 \implies {\pm \sqrt{-4\omega_0^2} \over 2}$$

and the general solution must be

$$y(t) = C_1\cos{w_0 t} + C_2\sin{w_0 t}$$

Finally $y$ can be rewritten as a single cosine function using the following relation

$$y(t) = R\cos{(\omega_0 t - \delta)}$$

where

$$\begin{align} R &= \sqrt{C_1^2 + C_2^2} \\ \delta &= \arctan{\frac{C_2}{C_1}}\end{align}$$

We see $y$ has the following properties:

• Amplitude of $R$ such that $y(t) \in [-R, R]$
• Periodic motion with period $T_0 = \frac{2\pi}{\omega_0}$
• Phase angle of $\delta$
• Frequency of $\omega_0 = \sqrt{\frac{k}{m}}$

Forced free vibrations and the resonance case

Adding a forcing function $f(t)$ to the first case gives us

$$x'' + \omega_0^2 x = \frac{f(t)}{m}$$

If the force applied $f(t) = F_0 \cos{\omega t}$, there are two cases:

For the first case, consider the frequency of the external force $\omega \ne \omega_0$. Then every solution has the form

$$y(t) = C_1\cos{\omega_0 t} + C_2\sin{\omega_0 t} + \frac{F_0}{m(\omega_0^2 - \omega^2)} \cos{\omega t}$$

and is the sum of two periodic functions of different periods.

However, if the frequency of the external force $\omega = \omega_0$ then this is called the resonance case. The forcing term $F_0\cos{\omega t}$ will cause unbounded oscillation; that is, an ever-increasing amplitude!

$$y(t) = C_1\cos{\omega_0 t} + C_2\sin{\omega_0 t} + \frac{F_0 t}{2m\omega_0} \sin{\omega_0 t}$$

Damped free vibrations

Including the effect of damping, the differential equation is

$$m x'' + 2\alpha x' + \omega_0^2x = 0$$

The characteristic polynomial will have roots:

$$r = \frac{-2\alpha \pm\sqrt{4\alpha^2 - 4m\omega_0^2}}{2m}$$

Let $g = 4\alpha^2 - 4m\omega_0^2$. Then if $g > 0$, both roots are negative and every solution has form

$$y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}$$

There is no oscillation and the system reaches equilibrium quickly. This case is called the overdamped case. A physical analogy could be the graph of capacitor discharging.

If $g = 0$, then every solution will be of form

$$y(t) = (C_1 + C_2t)e^{-\omega_0^2 t / 2m}$$

and there is still no oscillation. The system reaches equilibrium extremely quickly. This case is called the critically damped case.

Finally, if $g < 0$, every solution is of form

$$y(t) = e^{-\omega_0^2 t / 2m}[C_1 \cos{\mu t} + C_2 \sin{\mu t}],\quad\mu = \sqrt{4\omega_0^2 m - 4\alpha^2}{2m}$$

This is the underdamped case. We can analyse it by rewriting it as before into the single cosine function

$$y(t) = Re^{-ct/2m}\cos{\mu t - \delta}$$

The displacement $y$ oscillates between the curves $y=\pm Re^{-ct/2m}$. The function is therefore a cosine curve with decreasing amplitude. The function can also be related to the Laplace transform.

The Laplace transform

{% definition(ref="Generating function") %} A representation of an infinite sequence of numbers as the coefficients of a formal power series.

References