Linear differential equations: Introduction

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Let \(q(x), p_0(x), p_1(x), \cdots, p_{n-1}(x)\) be functions of \(x\). The equation

\[y^{(n)} + p_{n-1}(x)y^{(n-1)} + \cdots + p_1(x)y' + p_0(x)y + q(x) = 0\tag{Eq:linode}\]

of an unknown function \(y = y(x)\) is called an \(n\)-th order linear differential equation. If \(q(x) = 0\), then (Eq:linode) is said to be homogeneous; otherwise, it is said to be inhomogeneous.



Let's rewrite (Eq:linode) using differential operators. Let \(D = \frac{d}{dx}\) denote the differential operator with respect to \(x\). That is, \(Dy = \frac{d}{dx}y = y'\) and \(D^ny = \frac{d^n}{dx^n}y = y^{(n)}\), etc. By combining these operators, we can define a new operator \(E\) by

\[E = D^n + p_{n-1}(x)D^{n-1} + \cdots + p_1(x)D + p_0(x). \tag{eq:diffop}\]

Then, (Eq:linode) is concisely denoted as

\[Ey + q(x) = 0.\]

Theorem (Linear combinations of solutions of a homogeneous linear ODE)

Let \(y_1(x)\) and \(y_2(x)\) be functions. For any \(a, b \in \mathbb{R}\), the following holds:

\[E(ay_1 + by_2) = aEy_1 + bEy_2\]

where \(E\) is the differential operator defined by (Eq:diffop). In particular, if \(y=y_1(x)\) and \(y = y_2(x)\) are the solutions of the homogeneous linear differential equation \(Ey = 0\), then \(ay_1(x) + by_2(x)\) is also a solution.

Proof. Let \(y = ay_1 + by_2\). For \(k \geq 0\), we have

\[D^ky = y^{(k)} = ay_1^{(k)} + by_2^{(k)}.\]

Thus,

\[\begin{eqnarray*} Ey &=& D^ny + p_{n-1}(x)D^{n-1}y + \cdots + p_1(x)Dy + p_0(x)y\\ &=& a\{D^ny_1 + p_{n-1}(x)D^{n-1}y_1 + \cdots + p_1(x)Dy_1 + p_0(x)y_1\}\\ && + b\{D^ny_2 + p_{n-1}(x)D^{n-1}y_2 + \cdots + p_1(x)Dy_2 + p_0(x)y_2\}\\ &=& aEy_1 + bEy_2. \end{eqnarray*}\]

■

This theorem indicates two things:

  1. The operator \(E\) is linear.
  2. The solutions of \(Ey = 0\) are closed under scalar (real numbers) multiplication and addition. In other words, the set of solutions of a linear ODE comprises a vector space. (This means that each solution can be regarded as a vector.)

Example. Consider 
\[y'' + \omega^2y = 0.\]
\(y_1 = \cos (\omega x)\) and \(y_2 = \sin(\omega x)\) are solutions of this ODE. In fact, \(y_1'' = -\omega^2\cos(\omega x) = -\omega^2y_1\) and \(y_2'' = -\omega^2\sin(\omega x) = -\omega^2 y_2\). For any \(a, b\in\mathbb{R}\), we have
\[(a\cos(\omega x) + b\sin(\omega x))'' = -\omega^2(a\cos(\omega x)+ b\sin(\omega x)).\]
Thus, \(y = a y_1 + by_2\) is also a solution. □

Theorem (General solution of an inhomogeneous linear ODE)

Consider an inhomogeneous differential equation:
\[Ey + q(x) = 0\]
and the corresponding homogeneous differential equation
\[Ey = 0.\]
Let \(y = y_0(x)\) be a special solution of \(Ey + q(x) = 0\). Then, the general solution of \(Ey + q(x) = 0\) is given by
\[y = y_0(x) + Y(x)\]
where \(Y(x)\) is the general solution of \(Ey = 0\).
Proof. Let \(y = y_0(x) + Y(x)\). Then, since \(Ey_0(x) + q(x) = 0\) and \(EY(x) = 0\), 
\[\begin{eqnarray}Ey +q(x) &=& \{Ey_0(x) + EY(x)\} + q(x)\\ &=& \{Ey_0(x) + q(x)\} + EY(x) = 0. \end{eqnarray}\]
Suppose that \(f(x)\) is an arbitrary solution of \(Ey + q(x) = 0\). Let \(Y(x) = f(x) - y_0(x)\). Then, \(f(x) = y_0(x) + Y(x)\). But
\[EY = Ey - Ey_0 = -q(x) - (-q(x)) = 0.\]
Thus, \(y = Y(x)\) is a solution of \(Ey = 0\). ■

Example. Consider the inhomogeneous linear ODE
\[y' + 2y - 3e^{4x} = 0.\]
\(y_0 = \frac{1}{2}e^{4x}\) is a special solution. In fact,
\[\left(\frac{1}{2}e^{4x}\right)' + 2\left(\frac{1}{2}e^{4x}\right) - 3e^{4x} = 0.\]
The corresponding homogeneous ODE
\[y' + 2y = 0\]
has a solution \(Y = e^{-2x}\). In fact, 
\[(e^{-2x})' + 2e^{-2x} = 0.\]
Thus, the general solution is
\[y = \frac{1}{2}e^{4x} + Ce^{-2x}\]
where \(C\) is a constant. □

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