The Inverse-D Recipe

We are interested in solving equations of the form

$(a \mathcal D^2 + b \mathcal D + c)y = Q(x).$

The case when $Q = 0$ is straightforward and its general solution is called the complementary function. The case when $Q \neq 0$ can be solved using judiciously chosen Ansatz, or as I would like to call it, educated trial-and-error. Nevertheless, if we can find at least one particular integral $y_{\mathrm P}$ such that

$(a \mathcal D^2 + b \mathcal D + c)y_{\mathrm P} = Q(x),$

then the general solution is given by $y = y_{\mathrm P} + y_{\mathrm C}$ . We last asked the question: Can we find $y_{\mathrm P}$ systematically, rather than in an ad-hoc manner?

The key idea is some experimentation and a notational shift. If we took the “inverse” of the operator $p(\mathcal D):= a \mathcal D^2 + b \mathcal D + c$ , we will obtain

$\displaystyle y = \frac{1}{p(\mathcal D)}(Q(x)).$

Strictly speaking, the operator $p(\mathcal D)$ does not have an inverse, since for any complementary function $y_{\mathrm C} \neq 0$ ,

$\begin{aligned} p(\mathcal D)(y_{\mathrm C}) &= 0 = p(\mathcal D)(0).\end{aligned}$

Yet, further technical tools in linear algebra allows us to effectively “ignore” $y_{\mathrm C}$ and treat $p(\mathcal D)$ as having an inverse with technical caveats. We shall not discuss them here, but simply agree that

$\displaystyle \frac{1}{p(\mathcal D)}(0) = 0.$

With that, we can formally define the inverse- $\mathcal D$ operator.

Definition 1. We will write $\displaystyle q(x) := \frac{1}{p(\mathcal D)}(Q(x))$ to mean the unique function (with caveats) such that $p(\mathcal D)(q(x)) = Q(x)$ . Abbreviated,

$\displaystyle q(x) := \frac{1}{p(\mathcal D)}(Q(x)) \quad \iff \quad p(\mathcal D)(q(x)) = Q(x).$

This allows us to write, without ambiguity,

$\displaystyle \begin{aligned} \frac{1}{p(\mathcal D)}(p(\mathcal D)(q(x))) &= q(x), \\ p(\mathcal D)\left( \frac{1}{p(\mathcal D)}(Q(x)) \right) &= Q(x). \end{aligned}$

As with many calculus-related properties, the inverse- $\mathcal D$ operator is linear.

Theorem 1. For any $Q_1,Q_2$ and real constant $c$ ,

$\displaystyle \begin{aligned}\frac{1}{p(\mathcal D)}(Q_1(x) + Q_2(x)) &= \frac{1}{p(\mathcal D)}(Q_1(x)) + \frac{1}{p(\mathcal D)}(Q_2(x)),\\ \frac{1}{p(\mathcal D)} (c \cdot Q_1(x)) &= c \cdot \frac{1}{p(\mathcal D)}(Q_1(x)).\end{aligned}$

Proof. This is a consequence of linear algebraic ideas. Nevertheless, we provide a proof here. For each $i$ , find $q_i(x)$ such that $p(\mathcal D)(q_i(x)) = Q_i(x)$ . Then

$\displaystyle \begin{aligned}p(\mathcal D)\left( \frac{1}{p(\mathcal D)}(Q_1(x)) + \frac{1}{p(\mathcal D)}(Q_2(x)) \right) &= p(\mathcal D)(q_1(x) + q_2(x)) \\ &= p(\mathcal D)(q_1(x)) + p(\mathcal D)(q_2(x)) \\ &= Q_1(x) + Q_2(x).\end{aligned}$

The proof of the second property is similar.

If $p(\mathcal D) = \mathcal D$ , then

$\displaystyle \frac{1}{\mathcal D}(Q(x)) = q(x)\quad \iff \quad \mathcal D(q(x)) = Q(x).$

The calculation that reverses differentiation is integration, and by convention, we will say that

$\displaystyle \int0\, \mathrm dx = \frac{1}{\mathcal D}(0) = 0.$

Indeed, the arbitrary constants in the context of differential equations turns out to get “absorbed” into $y_{\mathrm C}$ .

Theorem 2. For any $Q(x) \neq 0$ , $\displaystyle \frac{1}{\mathcal D}(Q(x)) = \int Q(x)\, \mathrm dx$ .

Now we need to know several common $Q(x)$ and their results after being plugged into the inverse- $\mathcal D$ operator.

Theorem 3. Let $p$ be a polynomial. We have the following inverse- $\mathcal D$ calculations. with $p(0) \neq 0$ . For any constant $k$ ,

if $p(0) \neq 0$ , then $\displaystyle \frac{1}{p(\mathcal D)}(k) = \frac{1}{p(0)}\cdot k$ ,
if $p(k) \neq 0$ , then $\displaystyle \frac{1}{p(\mathcal D)}(e^{kx}) = \frac{1}{p(k)}\cdot e^{kx}$ ,
if $p(-k^2) \neq 0$ , then $\displaystyle \frac{1}{p(\mathcal D^2)}(\sin kx) = \frac{1}{p(-k^2)}\cdot \sin kx$ ,
if $p(-k^2) \neq 0$ , then $\displaystyle \frac{1}{p(\mathcal D^2)}(\cos kx) = \frac{1}{p(-k^2)}\cdot \cos kx$ .

Proof. We will prove the first result and leave the rest as exercises. Let $p(x) = a_0 + a_1x + \cdots + a_nx^n$ . We notice that for any constant $k$ , if $j \geq 1$ , then $\mathcal D^j(k) = 0$ . Hence,

$p(\mathcal D)(k) = a_0 k + a_1 \mathcal D(k) + \cdots + a_n \mathcal D^n(k) = a_0k = p(0)\cdot k.$

Hence, by the linearity of inverse- $\mathcal D$ operators,

$\displaystyle k = \frac{1}{p(\mathcal D)}(p(0) \cdot k) = p(0) \cdot \frac{1}{p(\mathcal D)}(k).$

Dividing both sides by $p(0) \neq 0$ yields the desired result.

Sadly, there are instances when $p(k) = 0$ , in which we need to reconsider our strategy. To that end, consider any function $W(x)$ . Then the product rule yields

$\displaystyle \begin{aligned}\mathcal D(e^{kx} W(x)) &= \mathcal D(e^{kx}) \cdot W(x) + e^{kx} \cdot \mathcal D(W(x))\\ &= ke^{kx} W(x) + e^{kx} \cdot \mathcal D(W(x))\\ &= e^{kx} (\mathcal D + k)(W(x)). \end{aligned}$

If we differentiate a second time, applying the same result but replacing $W(x)$ with $(\mathcal D + k)W(x)$ yields

$\displaystyle \begin{aligned}\mathcal D^2(e^{kx} W(x)) &= \mathcal D(e^{kx} (\mathcal D + k)W(x)) = e^{kx} (\mathcal D + k)^2(W(x)). \end{aligned}$

Repeating the pattern, we see that

$\displaystyle \begin{aligned}\mathcal D^n (e^{kx} W(x)) &=e^{kx} (\mathcal D + k)^n (W(x)). \end{aligned}$

By multiplying by $a_n$ and summing relevant terms, we obtain the intriguing result

$p(\mathcal D) (e^{kx} W(x)) = e^{kx} p(\mathcal D + k) (W(x)).$

But how do we translate this information into inverse- $\mathcal D$ operators? We perform a clever trick.

Theorem 4. Let $p$ be a polynomial. For any constant $k$ and any function $V(x)$ ,

$\displaystyle \frac 1{p(\mathcal D)} (e^{kx} V(x)) = e^{kx} \cdot \frac{1}{p(\mathcal D + k)}(V(x)).$

Proof. Define $\displaystyle W(x) = \frac{1}{p(\mathcal D+k)}(V(x)) \iff V(x) = p(\mathcal D+k)(W(x))$ . Then

$\displaystyle \begin{aligned} p(\mathcal D) \left( e^{kx} \frac{1}{p(\mathcal D+k)}(V(x)) \right) &= p(\mathcal D) (e^{kx} W(x)) \\ &= e^{kx} p(\mathcal D + k) (W(x)) \\ &= e^{kx} p(\mathcal D + k) \left( \frac{1}{p(\mathcal D+k)}(V(x)) \right) \\ &= e^{kx}(V(x)). \end{aligned}$

Taking $\displaystyle \frac 1{p(\mathcal D)}$ on both sides yields the desired result.

A common setup involves $Q(x)$ being a trigonometric function, but in that instance, what if $p(-k^2) = 0$ ? By the factor theorem, it turns out that this must mean that $\mathcal D^2 + k^2$ is a factor of $p(\mathcal D^2)$ . Hence, we study the latter case for simplicity.

Just like in Theorem 4, we consider any function $W(x)$ . Then some differentiation (left as an exercise) yields

$\displaystyle \begin{aligned}\mathcal D^2(\sin kx \cdot W(x)) &= \mathcal D^2(\sin kx) \cdot W(x) + 2 \cdot \mathcal D(\sin kx) \cdot \mathcal D(W(x)) + \sin kx \cdot \mathcal D^2(W(x)) \\ &= -k^2\sin kx \cdot W(x) + 2k \cos kx \cdot \mathcal D(W(x)) + \sin kx \cdot \mathcal D^2(W(x)).\end{aligned}$

The first equality does resemble the identity $(a+b)^2 = a^2 + 2ab + b^2$ , and indeed, this is the generalised product rule. For the right-hand side, the clever choice $W(x) = x$ yields $\mathcal D(W(x)) = 1$ and $\mathcal D^2(W(x)) = 0$ . This choice of $W(x)$ yields

$\mathcal D^2(x \sin kx ) = -k^2 x \sin kx + 2k \cos kx.$

Doing some algebra,

$(\mathcal D^2 + k^2)(x \sin kx) = 2 k \cos kx.$

Finally, taking the inverse- $\mathcal D$ operator and applying linearity yields our final result. The result is analogous if we started with $\cos kx \cdot W(x)$ .

Theorem 5. For any constant $k$ ,

$\displaystyle \begin{aligned} \frac 1{\mathcal D^2 + k^2} (\sin kx) &= -\frac{x \cos kx}{2k},\\ \frac 1{\mathcal D^2 + k^2} (\cos kx) &= \frac{x \sin kx}{2k}. \end{aligned}$

For a somewhat more elegant proof, of some of these identities, see this post. Finally, to see that this inverse- $\mathcal D$ recipe really works, let’s revisit our tedious problem from a previous post.

Example 1. Find the general solution to the differential equation

$\displaystyle (\mathcal D^2 - 5\mathcal D + 6)(y) = e^{2x}.$

Solution. For the complementary function $y_{\mathrm C}$ , we solve the characteristic equation $m^2 - 5m + 6 = 0$ to obtain the real and distinct roots $m = 2,3$ , yielding $y_{\mathrm C} = C_1 e^{2x} + C_2 e^{3x}$ .

For the particular integral $y_{\mathrm P}$ , we apply the inverse- $\mathcal D$ operator to get

$\displaystyle y_{\mathrm P} = \frac{1}{\mathcal D^2 - 5\mathcal D + 6} (e^{2x}).$

If we substitute $\mathcal D$ with $2$ , we get $2^2 - 5\cdot 2 + 6 = 0$ , which is sad, since we cannot use Theorem 3. Instead, we will use Theorem 4 with $V(x) = 1$ to get

$\displaystyle \begin{aligned} y_{\mathrm P} = \frac{1}{\mathcal D^2 - 5\mathcal D + 6} (e^{2x} \cdot 1) &= e^{2x} \cdot \frac{1}{(\mathcal D + 2)^2 - 5(\mathcal D + 2) + 6} (1) \\ &= e^{2x} \cdot \frac{1}{\mathcal D^2 - \mathcal D} (1) \\ &= e^{2x} \cdot \frac{1}{\mathcal D(\mathcal D - 1)} (1) \\ &= e^{2x} \cdot \frac{1}{\mathcal D} \left( \frac{1}{\mathcal D-1}(1) \right). \end{aligned}$

We leave it as an exercise to verify that the last two steps are legitimate, i.e., in the sense that for polynomials $p, q$ ,

$\displaystyle \frac{1}{p(\mathcal D)q(\mathcal D)}(Q(x)) = \frac{1}{p(\mathcal D)}\left( \frac{1}{q(\mathcal D)} (Q(x)) \right) = \frac{1}{q(\mathcal D)}\left( \frac{1}{p(\mathcal D)} (Q(x)) \right).$

Back to the problem, applying results from Theorems 2 and 3 yield

$\displaystyle \begin{aligned} y_{\mathrm P} &= e^{2x} \cdot \frac{1}{\mathcal D} \left( \frac{1}{\mathcal D-1}(1) \right) = e^{2x} \cdot \frac{1}{\mathcal D} \left(\frac 1{0-1} \cdot 1\right) \\ &= -e^{2x} \cdot \frac{1}{\mathcal D} (1) = -e^{2x} \cdot \int 1\,\mathrm dx = -e^{2x} \cdot x = -xe^{2x}.\end{aligned}$

Hence, the desired general solution to the original differential equation must be

$y = y_{\mathrm P} + y_{\mathrm C} = -xe^{2x} + C_1 e^{2x} + C_2 e^{3x},$

which matches our previously obtained result.

—Joel Kindiak, 31 Jan 25, 1608H

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