Linear ODEs —
Operators, Superposition
& Fundamental Sets

What is a differential operator?

The symbol \(D\) is just shorthand for:

So \(D\) is an operator. An operator is just something that acts on a function and produces another function.

So if \(y(t) = t^3\), then:

So \(D\) means: "take one derivative." Then:

So \(\dfrac{d^3y}{dt^3} = D(D(Dy)) = D^3y\) — just compressed notation.

Mental Model

Think of \(D\) like a machine:

Then \(D^2\) means run the machine twice. \(D^3\) means run it three times. Old-school slick math move. Very clean.

What does it mean to combine differential operators into a polynomial expression?

Consider the general linear operator:

acting on \(y\), and equal to \(g(t)\). What this really means is:

So the operator \(L\) is just a compact way of writing a linear differential equation. For example, the ODE

can be written as

because \(D^2 y = y''\), \(3Dy = 3y'\), and \(-4y = -4y\). So:

That is why they call it a polynomial expression in \(D\). It behaves like algebra, except the variable is the derivative operator.

Why this matters

This is what leads to the characteristic equation trick for constant-coefficient linear ODEs:

Boom. That is where the algebra pipeline starts.

What is \(L\)?

For a second-order ODE, the linear differential operator \(L\) looks like:

So if you feed \(y\) into \(L\), it outputs:

That is just notation. Nothing mystical. It is like defining a function called \(L\) that takes in a function \(y\) and spits out a new expression involving \(y, y', y''\).

What is the principle of superposition?

This is the juicy part.

If \(y_1(t)\) and \(y_2(t)\) are solutions, then

is also a solution. This is true for a linear homogeneous differential equation. That word combo matters:

• linear — \(y, y', y''\) only appear to the first power, not multiplied together, no weird nonlinear stuff
• homogeneous — right side is \(0\)

Example: \(y'' - 3y' + 2y = 0\). If \(y_1 = e^t\) is a solution and \(y_2 = e^{2t}\) is a solution, then \(y = c_1 e^t + c_2 e^{2t}\) is also a solution.

Why?

Because derivatives distribute over addition and constants pull out nicely. Let's show it with no hand-waving.

Suppose \(L[y] = P(t)y'' + Q(t)y' + R(t)y\), and suppose:

Now take \(y = c_1 y_1 + c_2 y_2\). Then:

Now plug into \(L[y]\):

Distribute:

That is:

But \(L[y_1] = 0\) and \(L[y_2] = 0\), so:

So \(y\) is also a solution. That is superposition.

Street-level intuition

Why do they call this "fundamental sets of solutions"?

Because for a second-order linear homogeneous ODE, you need two linearly independent solutions to build the full general solution.

So if you find two good independent solutions \(y_1\) and \(y_2\), then they form a fundamental set of solutions, and the general solution is:

Not just any two solutions — they need to be linearly independent. That means one cannot just be a constant multiple of the other.

If they are dependent, then you really only have one direction, not two.

And for a second-order equation, you need two degrees of freedom because you usually need to satisfy two initial conditions, like:

So the constants \(c_1\) and \(c_2\) give you the flexibility to match those conditions.

Tiny caveat

The statement \(y(t) = c_1 y_1(t) + c_2 y_2(t)\) is definitely true if the DE is linear homogeneous.

If the equation is nonhomogeneous, like \(y'' + y = \sin t\), then superposition of two full solutions does not work the same way. For nonhomogeneous equations, the general solution is:

where:

• \(y_c\) — complementary solution to the homogeneous part
• \(y_p\) — one particular solution to the forcing

So superposition as stated is really about the homogeneous linear case.

What it all means in one clean chunk

Here is the unified meaning:

• Differentiation can be treated like an operator \(D\)
• Higher derivatives are powers of that operator: \(D^2 y, D^3 y, \dots\)
• A linear differential equation can be written compactly using operator notation
• For a linear homogeneous ODE, the operator \(L\) is linear
• Because \(L\) is linear, if \(y_1\) and \(y_2\) solve the equation, then any linear combination \(c_1 y_1 + c_2 y_2\) also solves it
• If \(y_1\) and \(y_2\) are linearly independent, they form a fundamental set
• Then the general solution is \(y = c_1 y_1 + c_2 y_2\)

Super quick concrete example

Take:

In operator form:

Two solutions are:

Then by superposition:

Check it:

Then:

Three Core Ideas

The three core ideas from this whole setup:

So this is really the foundation for why second-order homogeneous linear equations have general solutions built from two independent basis solutions. Pretty rad setup, honestly.