Lecture notes · Differential Equations · Lecture 24

Homogeneous Linear Systems with Constant Coefficients

MIT 18.03 Differential Equations, lecture 24, explained in plain English

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Part 1 · 0:00–10:00

This part shows how to write a pair of linked change equations as a matrix problem, which is the method used in practice, and it sets up a trial solution.

Key ideas

The lecturer reuses the egg-cooling system, but swaps the temperature names for neutral letters x and y so the work is more general. 0:39
A matrix is a grid of numbers. The two equations can be written as one matrix equation: a column vector of (x, y) with a derivative, equal to a grid of coefficients times (x, y). 4:13
The general solution is a sum of two column vectors. Each one is a constant times a scalar function, such as e to the negative t, and a column of numbers. 6:09
Constant factors go out front, the exponential function goes at the back, and the column of numbers sits in the middle. This layout is the standard way to write it. 6:32
The new method uses a trial solution, based on the same idea as the e to the rt guess for second-order equations. The lecturer says a simple guess with two different constants will not work. 8:07

Words to know

matrix: A neat grid of numbers in rows and columns. Here it holds the numbers that multiply x and y in the equations.
column vector: A stack of numbers or expressions in one column. Here (x, y) is stacked this way.
elimination: A method of solving equations by cancelling variables one at a time. The lecturer says it is mostly used by people who have forgotten other methods.
trial solution: A guessed answer with some unknown parts, which you then test in the equation. Here the guess has the form of a constant times e to the rt.

Example

Think of two cups of tea that cool by swapping heat with each other, so each one's change depends on the other. Writing both changes as one neat table of numbers is like writing a shared recipe once instead of two separate ones. The table is easier to read and to solve.

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

Part 2 · 10:00–20:00

This part shows how to turn a pair of constant-coefficient differential equations into algebra by guessing an exponential solution and then solving for the right numbers.

Key ideas

Use one trial solution with the same exponential factor, e to the lambda t, for both x and y, multiplied by two unknown numbers. Lambda is just a name for an unknown exponent. 10:14
Substituting the guess into the system makes the e to the lambda t part drop out. Calculus becomes algebra, which is the same idea used in the Laplace transform. 12:30
Treat lambda as a parameter, a fixed number that you are not solving for right now. Then the equations are linear and homogeneous in the two unknowns a1 and a2. 15:27
A square homogeneous system always has the trivial solution, all zeros. You want a nontrivial solution, which exists only when the determinant is zero. 17:59
Expanding the determinant gives a quadratic equation in lambda: lambda squared plus 7 lambda plus 6 equals zero. Those lambda values are the ones that lead to useful solutions. 19:21

Words to know

determinant: A single number calculated from a grid of numbers in a square system. For a 2 by 2 grid, multiply the diagonal pairs and subtract. If it is zero, the system has non-zero solutions.
parameter: A number that stays fixed while you solve for other unknowns. Here lambda is treated as fixed so that a1 and a2 are the only unknowns.
trivial solution: The boring answer where every unknown is zero. It always works for these systems but tells you nothing useful.
nontrivial: Not the boring all-zero answer. It means at least one unknown is not zero, which is the kind of answer you want.

Example

Imagine a savings account that grows at one fixed rate, with the same growth pattern for every account you check. Once you assume the growth has one shape, the hard calculus question becomes a simpler question about one number. That is the same move this lesson makes with lambda.

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Part 3 · 20:00–30:00

This part shows how to find the special lambda values, use each one to build a solution, and combine the solutions into the general answer.

Key ideas

The method turns the problem into a characteristic equation, a quadratic equation whose roots are the special lambda values. Factoring it gives lambda = -1 and lambda = -6. 20:34
Each lambda is a separate problem. For each one, you plug it into the system of equations and solve for the numbers a1 and a2. A correct lambda makes one equation a constant multiple of the other, so the system has a nonzero answer. 22:56
For a two-by-two system, you can pick a value for one variable, such as a2 = 1, and solve for the other. Any nonzero multiple of that answer also works. 24:54
Each solution is that (a1, a2) pair multiplied by e raised to the lambda times t. Because the system is linear, you can add the solutions with arbitrary constants to get the general solution. 27:54
The lecturer says this method gives the same answer as elimination, which is a useful check. The two methods take different routes but end up at the same result. 28:47

Words to know

characteristic equation: A simpler equation, built from the original problem, whose solutions (roots) tell you the special lambda values to try.
trivial solution: The answer where every number is zero. It always works but tells you nothing useful, so you look for a nonzero answer.
superposition principle: For linear problems, if two answers work, then any constant multiple of each one, or their sum, also works.
arbitrary constant: A letter standing for any number you like. It covers every possible value, so renaming or rescaling it does not change the answer.

Example

Think of a recipe that turns out right only at certain oven settings. You test each setting one at a time, and only the correct settings give a result that holds together. Here, the correct lambda values are like those oven settings, and each one gives its own working answer.

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

Part 4 · 30:00–40:00

This part shows how to find the special hidden numbers, called eigenvalues, of a 2-by-2 matrix, using a quick formula you should memorize.

Key ideas

When you substitute a trial solution into a homogeneous system, you get a condition: the system has a nontrivial (not all zero) solution only if the determinant of the coefficients is zero. 33:13
For a 2-by-2 matrix (a, b; c, d), the characteristic equation is lambda squared minus (a + d) lambda plus (ad minus bc) equals zero. You can write it down directly without redoing the derivation. 34:45
The trace is the sum of the diagonal elements of a square matrix. In the constant term, ad minus bc is the determinant, and a + d is the trace. 36:03
The roots of the characteristic equation are the eigenvalues. You cannot see them by looking at the matrix; you have to calculate them, but they control how the system behaves. 37:39

Words to know

eigenvalues: Special numbers that belong to a matrix. They are the roots of its characteristic equation, and they control how the system behaves. You must calculate them; they are not visible in the matrix.
characteristic equation: A quadratic equation built from a matrix. Its roots are the eigenvalues. For a 2-by-2 matrix you can write it straight from the entries.
determinant: For a 2-by-2 matrix (a, b; c, d), a single number equal to ad minus bc. The system has a nontrivial solution only when this is zero.
trace: The sum of the diagonal elements of a square matrix, running from top left to bottom right. For a 2-by-2 matrix it is a + d.

Example

The lesson compares this method to solving a second-order equation. For a second-order equation you write down its characteristic equation, factor it, find the roots, and build the solution. The matrix method works the same way: write the characteristic equation, find its roots (the eigenvalues), and use them to build the solution.

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

Part 5 · 40:00–48:59

This part shows how to find the eigenvectors that go with each eigenvalue, and how to write the general solution of a constant-coefficient system in a short matrix form, x' = Ax.

Key ideas

Once you have the eigenvalues, the last step is to find an eigenvector for each one. You do this by solving a system with a zero on the right side. 40:16
The general solution is built by adding together the eigenvectors, each multiplied by its own arbitrary constant and by e raised to its eigenvalue times t. This adding-up rule is called the superposition principle. 42:20
In matrix form, the system is written as x' = Ax. The same method works for any number of equations, not just two. Plugging in a trial solution, x = α e^(λt), and canceling the e^(λt) gives Aα = λα. 44:05
You cannot subtract a plain number from a matrix. So λ is put on the diagonal of an identity matrix (a matrix with 1s on the diagonal and 0s elsewhere), and the equation becomes (A − λI)α = 0. 46:39
The eigenvalues come from the characteristic equation, which says the determinant of (A − λI) is zero. The roots of that equation are the eigenvalues. 48:14

Words to know

eigenvalue: A special number tied to a matrix. In this lesson, each one gives an exponential piece of the solution, like e raised to λt.
eigenvector: A special vector that goes with one eigenvalue. It is found by solving (A − λI)α = 0 for that eigenvalue.
characteristic equation: The equation det(A − λI) = 0. Its roots are the eigenvalues, so solving it is the first step.
identity matrix: A square grid of numbers with 1s down the main diagonal and 0s everywhere else. Multiplying by it leaves a vector unchanged, so it lets you turn λ into a matrix of the right size.

Example

Think of a set of gears that all turn at their own steady pace. Each gear's pace is like an eigenvalue, and the way it moves is like its eigenvector. The full motion is the sum of all these separate movements, which is the superposition idea from this part.

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

← Lecture 22: Using Laplace Transform to Solve ODEs with Discontinuous Inputs

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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