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4. Operators

We now begin a close look at the linear case.

Definition 4.1 (First-order linear ODE)

A first-order linear ODE is an equation of the form

(4.1)\[\dd{x}{t} = a(t) x + f(t).\]

We call \(a(t)\) the coefficient function and \(f(t)\) the forcing function.

Attention

A linear ODE has a linear dependence on the unknown (dependent) variable \(x\). It may have arbitrary dependence on the independent variable \(t\). Also, the solution \(x\) is usually not a linear function of \(t\).

Warning

If presented with a linear ODE in the form \(b(t)x' + c(t)x = g(t)\), you should divide through by \(b\) and rearrange so that the equation is in standard form (4.1) if you want to apply one of our formulas.

There are some strong analogies between (4.1) and the linear algebraic system \(\bfA\bfx=\bfb\). Some new notation helps to clarify the similarities.

Definition 4.2 (Linear operator)

A linear operator \(\opA\) is a rule for transforming functions to other functions, such that

\[\begin{align*} \opA[cx(t)] & =c\opA[x(t)], \\ \opA[x(t) + y(t)] &= \opA[x(t)] + \opA[y(t)], \end{align*}\]

for all functions \(x,y\) and numbers \(c\).

Note

You may recognize the two conditions in this definition as restatements of the properties of a linear function.

In this context we are interested in the operator

(4.2)\[\opA[x] = x' - a(t)x,\]

whose linearity you can easily check for yourself against the definition. We can now express the ODE (4.1) simply as

\[ \opA[x]=f. \]

Example

The equation

\[ t x' = \sin(t) - x \]

is linear. To write it in operator form, we rearrange to

\[ x' + \frac{1}{t}x = \frac{\sin(t)}{t}. \]

Thus the linear operator for this ODE is \(\opA[x]=x' + \tfrac{1}{t}x\), and the ODE is

\[ \opA[x] = \frac{\sin(t)}{t}. \]

4.1. Homogeneous solutions

As with linear systems, we have a special role for the homogeneous linear ODE \(\opA[x]=0\), or

\[ \dd{x}{t} = a(t)x, \]

which is (4.1) with zero forcing.

Theorem 4.3 (Superposition)

If \(x_1(t), x_2(t),\ldots x_k(t)\) are solutions of \(\opA[x]=0\), then so is any linear combination \(c_1x_1 + \cdots + c_kx_k\) for constants \(c_j\).

Proof

Because of linearity we can write

\[\opA[c_1x_1 + \cdots + c_kx_k] = c_1\opA[x_1] + \cdots + c_k\opA[x_k].\]

By assumption, each \(\opA[x_j]=0\). So the sum on the right is zero as well.

4.2. General solutions

For the problem with nonzero forcing, we can follow the same script as for general solutions of \(\bfA\bfx=\bfb\) to get the following result.

Theorem 4.4

All solutions of \(\opA[x]=f\) may be written as

\[x(t) = x_h(t) + x_p(t),\]

where \(x_h\) is the general solution of \(\opA[x]=0\) and \(x_p\) is any particular solution of \(\opA[x]=f\). We call this the general solution of the linear ODE.

We have arrived at a solution strategy for \(\opA[x]=f\).

Algorithm 4.5 (Solution of a first-order linear ODE)

  1. Find the general solution \(x_h\) of the associated homogeneous problem \(\opA[x]=0\).

  2. Find any particular solution \(x_p\) of the original \(\opA[x]=f\).

  3. Add them.

  4. If an initial condition is given, solve for the integration constant.

We elaborate on steps 1 and 2 in the next several sections.

3. Initial-value problems 5. Homogeneous solutions