Lecture notes · Differential Equations · Lecture 9

Solving Second-order Linear ODE's with Constant Coefficients

MIT 18.03 Differential Equations, lecture 9, explained in plain English

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Part 1 · 0:00–10:00

This part introduces a basic type of second-order equation, explains the shape its general solution should take, and starts a physical example about a spring, a mass, and a dashpot.

Key ideas

The lecturer studies the simplest case, where the right-hand side is zero. This is called homogeneous, and the lecture starts here before moving to harder cases. 1:33
A second-order equation has a general solution with two arbitrary constants, c1 and c2. The number of constants matches the order of the equation. 3:56
Solving the equation comes down to finding two solutions, y1 and y2. The general solution is built by multiplying each one by a constant and adding them together. 4:53
If you have initial conditions, which are starting values for the problem, you pick c1 and c2 so the solution matches them. 5:49

Words to know

homogeneous: An equation is homogeneous when the right-hand side is zero, meaning nothing outside the equation is pushing on it.
arbitrary constants: Unknown numbers that can be any value. They stay in the solution until you pick values for them using extra information.
dashpot: A device with a piston in a chamber of air that slows down motion. It is the same kind of part that stops a door from slamming shut.
initial conditions: The starting values of a problem, such as where something is or how fast it is moving at the beginning.

Example

A door closer on a door is a dashpot. It slows the door so it does not slam shut, and the air pushed through the chamber is what does the slowing. The lecturer uses this same kind of device in a spring and mass example.

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Part 2 · 10:00–20:00

This part shows how to find the solutions to a spring-mass-damper equation by guessing an exponential form and turning the problem into a quadratic equation.

Key ideas

To solve the equation, you need two solutions that are independent, meaning one is not just a constant multiple of the other. Two independent solutions let you build the full general answer. 12:41
The basic method is to guess that the answer has the form e to the rt, where r is a number you still need to find. Only the constant in the exponent can be changed, so that is the value to pick. 13:41
When you put this guess into the equation, every term has e to the rt as a factor. Since e to the rt is never zero, you can divide it out, leaving a simple quadratic equation in r called the characteristic equation. 15:39
The roots of the characteristic equation fall into three cases: real and unequal, a pair of complex conjugate numbers, or real and equal. Each case gives a different kind of motion. 17:02
In the real and unequal case, the two roots give two solutions right away, and the general solution is c1 times e to the r1 t plus c2 times e to the r2 t. The constants c1 and c2 are set later using starting conditions. 18:19

Words to know

dashpot: A shock absorber part that pushes back against motion. Its force is always opposite to the direction the mass is moving.
independent: Two solutions are independent when they look different, so neither one is just a fixed multiple of the other.
exponential: A function like e to the rt, where the variable t sits in the power. It grows or shrinks very fast, and it is easy to differentiate.
characteristic equation: The simple quadratic equation, r squared plus a times r plus b equals zero, that you get from the original equation. Solving it tells you how the system behaves.

Example

Think of a car's suspension: a spring holds the car up and a shock absorber slows down its bouncing. The shock absorber pushes back only when the car moves, which is like the dashpot resisting velocity. The lesson uses this same spring-mass-dashpot picture to make the equation easier to picture.

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Part 3 · 20:00–30:00

This part shows how to solve a spring-and-damper equation step by step, and how the shape of the answer depends on whether the roots are real or complex.

Key ideas

Write the characteristic equation first. For this example, r squared plus 4r plus 3 equals zero, and factoring gives roots of minus 3 and minus 1. 20:17
The general solution is a mix of two exponential terms, each with its own constant. Because there are two constants, you need two conditions to find them. 21:53
The starting position and starting speed (initial conditions) give two simple equations with two unknowns. Solving them gives c1 equal to minus one half and c2 equal to three halves. 23:49
When the damping is large, the mass slowly returns to the equilibrium position without crossing it. This is called the over-damped case. 27:05
When the damping is small, the roots are complex, written as a plus or minus bi, and the mass oscillates. A theorem lets you take the real and imaginary parts of a complex solution to get real solutions. 29:26

Words to know

characteristic equation: A simple equation made from the original problem, using r in place of the derivatives. Its solutions tell you what the exponential terms look like.
roots: The answers to the characteristic equation. They are the numbers that set the exponents in the solution.
over-damped: Damping so strong that the moving part slowly settles back without swinging past its resting point.
complex conjugate: A pair of numbers of the form a plus bi and a minus bi, where i is the square root of minus one. They appear together as roots.

Example

Think of a screen door with a slow closer. If the closer is strong, the door eases shut without bouncing or swinging past the frame. If the closer is weak, the door swings back and forth several times before it stops.

3 checked practice questions for this part are in the app.

Part 4 · 30:00–40:00

This part shows how to turn a complex-number answer into two real solutions for a spring-and-damping equation, then works an example with a stiff spring.

Key ideas

If you find a complex solution to an equation with real coefficients, its real part and its imaginary part are each a real solution. Those two parts give you the two solutions you need. 30:20
This only works because the coefficients are real numbers. If a coefficient had an imaginary piece, you could not cleanly split the answer into a real part and an imaginary part. 30:39
For the complex case, the real and imaginary parts of e to the power (a plus b i) t are e^(at) times cos(bt) and e^(at) times sin(bt). Both are solutions. 33:37
The general solution is e^(at) times (c1 cos bt + c2 sin bt). The e^(at) part controls how the size of the swing changes over time, and the cos and sin part makes a steady back-and-forth wiggle. 34:32
In the example, a stiffer spring with weak damping gives roots of minus 2 plus or minus i. The e^(-2t) factor shrinks the swing over time, and the leftover cos t plus 2 sin t combines into one wave with amplitude square root of 5. 39:24

Words to know

real part: The part of a complex number that has no 'i' attached. It is an ordinary number you could put on a regular number line.
imaginary part: The part of a complex number that is multiplied by 'i'. The lesson uses it to split a complex answer into two separate answers.
characteristic equation: A simpler equation made by swapping the derivatives in the original equation for a letter r. Solving it tells you the exponential pieces of the answer.
amplitude: How far a wave swings away from its middle position. A bigger amplitude means a bigger swing.

Example

Think of a child on a swing that someone pushes once and then lets go. The swing goes back and forth at a steady rate while its swing slowly gets smaller, because of air resistance. That is like the e^(-2t) factor shrinking the wiggle while the cos and sin parts keep it moving back and forth.

3 checked practice questions for this part are in the app.

Part 5 · 40:00–49:56

This part explains the in-between damping case, called critically damped, and how to find the second solution it needs.

Key ideas

In the underdamped case, a spring-and-damper system swings back and forth, and in theory the swinging never fully stops. 41:54
Critically damped is the just-right case that sits between the swinging case and the other case, so motion stops without repeated swinging. Engineers aim for this when they want to calm unwanted motion. 42:16
The critically damped case happens when the characteristic equation has two equal roots, which gives one root, -a, repeated twice. 43:40
With a double root, e to the minus a t is only one solution, so a second, different solution must be found. The lecture shows a method: multiply by a function u, which leads to u double prime equals zero. 44:39
Solving u double prime equals zero gives u equals c1 t plus c2, so the second solution is t times e to the minus a t. 49:06

Words to know

critically damped: The exact amount of damping (slowing) that stops a system from swinging, sitting between the swinging case and the slow case.
characteristic equation: A simpler equation made from the original equation by swapping each derivative for a power of r. Its roots tell you the form of the solutions.
double root: A root that appears twice, meaning two roots are equal to each other.
damping: Something that slows down or drains energy from a moving system, like a door closer or a shock absorber.

Example

Think of a door on a spring. If it has too little damping, it swings back and forth many times. With just the right damping, it closes smoothly and stops without bouncing back.

3 checked practice questions for this part are in the app.

← Lecture 3: Solving First-order Linear ODEs · Lecture 10: Continuation: Complex Characteristic Roots →

Notes written by AI (Claude) from the lecture's official transcript, so check anything important against the lecture. MIT OpenCourseWare · CC BY-NC-SA 4.0 · not affiliated with or endorsed by MIT. These notes are shared under the same license.

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