Constant-Coefficient Equations

Second-order linear equations with constant coefficients are very important, especially for applications in mechanical and electrical engineering (as we will see). The general second-order constant-coefficient linear equation is $T y = y'' + p y' + q y = g(x)$, where $p$ and $q$ are constants. We will be especially interested in the cases where either $g(x) = 0$ (the homogeneous case) or $g(x) = \exp(a x)$ for some constant $a$.

The main idea in solving these equations is: make an educated guess at the form of the solution, and see what has to happen to make a function of this form satisfy the equation. In the homogeneous case, it turns out that the form to use is $y = \exp(r x)$, where $r$ is a constant. If we plug this in to the differential equation, we get $T y = r^2 \exp(r x) + p r \exp(r x) + q \exp(r x) = 0$, and factoring out the $\exp(r x)$ (which is not $0$) we are left with

$$r^2 + p r + q = 0$$

This is called the characteristic equation of the operator $T$ (or of the differential equation). If $r$ is a root of the characteristic equation, $y = \exp(r x)$ is a solution of the differential equation.

Recall that we are looking for a fundamental set consisting of two linearly independent solutions. Now a quadratic equation may have two real roots, and this would give us our two solutions (it is not hard to see that they are linearly independent).

Example: Consider the equation $y'' + 2 y' - 3 y = 0$. The characteristic equation $r^2 + 2 r - 3 = 0$ has two roots $r = -3$ and $r = 1$. Therefore $\exp(-3 x)$ and $\exp(x)$ form the fundamental set of solutions. The general solution is

$$y = c_1 \exp(-3 x) + c_2 \exp(x)$$

Now, a quadratic equation may also have only one real root, or no real roots. We will have to deal with these possibilities as well.

At this point it is useful to introduce the differentiation operator $D$ defined by $Dy = y'$. We can then write the operator $Ty = y'' + p y' + q y$ of our differential equation as $T = D^2 + p D + q$, a polynomial in $D$. We can manipulate polynomials in $D$ just as we would manipulate ordinary polynomials. For example, we might factor $D^2 + 2 D - 3 = (D-1)(D+3)$.

What does $(D-1)(D+3)$ mean? These are operators, so they are defined by what they do to functions. The ``multiplication'' of operators is really composition: you first let the operator on the right act on the function, and then the operator on the left acts on the result. Thus for $(D-1)(D+3)y$ you first take $(D+3)y = y' + 3y$, then $(D-1)(y' + 3 y) = D(y' + 3 y) - (y' + 3 y) = y'' + 2 y' - 3 y$.

The characteristic polynomial is related to the way the $D$ operator acts on exponential functions: $D \exp(r x) = r \exp(r x)$. So, for example, $(D-1)(D+3) \exp(r x) = (D-1)(r+3)\exp(r x) = (r-1)(r+3)\exp(r x)$. Or in general, for any polynomial $P$,

$$P(D) \exp(r x) = P(r) \exp(r x)$$

This has the following consequences:

• If $P(r) = 0$ then $\exp(r x)$ is a solution of the homogeneous equation $P(D) y = 0$.
• If $P(a) \ne 0$ then one solution of the non-homogeneous equation $P(D) y = \exp(a x)$ is $y = \exp(a x)/P(a)$.

Now how can we deal with the homogeneous equation $P(D) y = 0$ when $P(r)$ has only one real root, or with the non-homogeneous equation $P(D) y = \exp(r x)$ where $P(r) = 0$? The key is the

Exponential Shift Theorem: For any polynomial $P$, constant $k$ and function $u$,

$$P(D) (\exp(k x) u) = \exp(k x) P(D + k) u$$

Proof: Note that $D (\exp(k x) u) = \exp(k x) u' + k \exp(k x) u = \exp(k x)(D+k) u$. Repeated application of this gives us the result for any power of $D$, e.g.

$$D^2(\exp(k x) u) = D(D(\exp(k x) u) = D (\exp(k x) (D+k) u)$$

$$\qquad = \exp(k x)(D+k)((D+k)u) = \exp(kx)(D+k)^2 u$$

Adding constants times powers of $D$ gives us all polynomials.

Now let's apply this to $P(D) y = 0$ where the polynomial $P(D)$ has only one root. This means $P(r) = (r-a)^2$ where $a$ is the root. We look for a solution of the form $y = \exp(a x) u$ for some function $u$. The Exponential Shift Theorem says $(D-a)^2 y = \exp(a x) D^2 u$, so we want $u'' = 0$. Two linearly independent solutions of this are $u= 1$ and $u = x$. Thus a fundamental set of solutions of $(D-a)^2 y = 0$ is $\exp(a x)$ and $x \exp(a x)$.

Similarly, we can try $P(D) y = \exp(a x)$ where $P(a) = 0$. Again we try $y = \exp(a x) u$, and get $P(D) y = \exp(a x) P(D+a) u$, so we want $P(D+a) u = 1$.

• If $P(D) = (D-b)(D-a)$ with $b \ne a$, $P(D+a)u = (D+a-b)D u = 1$. Writing $v = Du$ we want $(D+a-b) v = 1$. Clearly one solution of this is the constant $v = 1/(a-b)$. Then from $Du=v$ we get a solution $u = x/(a-b)$, i.e. $y = x \exp(a x)/(a-b)$.
• If $P(D) = (D-a)^2$, $P(D+a) u = D^2 u = 1$. Clearly one solution is $u = x^2/2$, i.e. $y = x^2 \exp(a x)/2$.

Example: Solve the initial value problem

$$y'' + 2 y' + y = 4 {\rm e}^{-x},\quad y(0) = 2,\quad y'(0) = 1$$

We have $P(D) = D^2 + 2 D + 1 = (D+1)^2$, with one root $r = -1$. Since the right side of the equation involves $\exp(r x)$, we use the Exponential Shift Theorem with $y = \exp(-x) u$. We have $P(D-1) u = D^2 u = 4$, so one solution is $u = 2 x^2$. This gives us one solution of our differential equation: $y_p = 2 x^2 \exp(-x)$. The fundamental set of solutions of the homogeneous equation $(D+1)^2 y = 0$ is $\exp(-x)$ and $x \exp(-x)$. Thus the general solution is $y = (2 x^2 + c_1 x + c_2) \exp(-x)$. From the initial conditions we have $y(0) = c_2 = 2$ and $y'(0) = c_1 - c_2 = 1$. Thus $c_1 = 3$ and $c_2 = 2$, and the answer is

$$y = (2 x^2 + 3 x + 2) {\rm e}^{-x}$$