KindiakMath

Category: Differential Equations

Radioactive Decay

Big Idea

To solve a separable differential equation

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = f(x)g(y),$

we separate the variables and integrate on each side:

$\displaystyle \int \frac 1{g(y)}\, \mathrm dy = \int f(x)\, \mathrm dx.$

This is called the method of separable variables. Remember to include a $+C$ on at least one side of the result! You may leave your answer in implicit form if there is no $|y|$ in your expression.

Questions

Question 1. Solve the differential equation

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = \frac{x}{\cos(y) + 1}.$

(Click for Solution)

Solution. By the method of separable variables,

$\begin{aligned} \frac{\mathrm dy}{\mathrm dx} &= \frac{x}{\cos(y) + 1} \\\int (\cos(y)+1)\ \mathrm dy &= \int x\ \mathrm dx \\ \sin(y) + y &= \frac {x^2}{2} + C. \end{aligned}$

Question 2. Solve the differential equation

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = \frac{1 + \cos(x)}{e^y}.$

(Click for Solution)

Solution. By the method of separable variables,

$\begin{aligned} \frac{\mathrm dy}{\mathrm dx} &= \frac{1 + \cos(x)}{e^y} \\\int e^y \ \mathrm dy &= \int (1 + \cos(x))\ \mathrm dx \\ e^y &= x + \sin(x) + C. \end{aligned}$

Question 3. A radioactive substance decays with a decay rate of $\lambda > 0$ . Let $x \equiv x(t)$ denote the amount of a radioactive substance at time $t$ . The radioactive decay model states that $x$ decreases at a rate proportional to $x$ with proportionality constant $\lambda$ .

Compute the half-life of the substance (i.e. the time taken for the substance to decay to half of its quantity) in terms of $\lambda$ .

(Click for Solution)

Solution. The rate of change of $x$ is $\displaystyle \frac{\mathrm dx}{\mathrm dt}$ . Since $\displaystyle \frac{\mathrm dx}{\mathrm dt}$ is proportional to $x$ and decreasing,

$\displaystyle \frac{\mathrm dx}{\mathrm dt} = -\lambda x.$

By the method of separable variables,

$\begin{aligned} \frac{\mathrm dx}{\mathrm dt} &= -\lambda x \\ \int \frac 1x\, \mathrm dx &= \int -\lambda\ \mathrm dt \\ \ln |x| &= -\lambda t + C \\ |x| &= e^{-\lambda t + C} \\ x &= \pm e^{-\lambda t} \cdot e^C \\ &= Ae^{-\lambda t}, \end{aligned}$

where $A := \pm e^C$ is an arbitrary constant. Now, $x(0) = A > 0$ . Since we want to find $t$ such that $x(t) = A/2$ , we solve the equation

$\displaystyle \begin{aligned} \frac A2 &= Ae^{-\lambda t} \\ \frac 12 &= e^{-\lambda t} \\ \ln \left(\frac 12 \right) &= -\lambda t \\ -\ln 2 &= - \lambda t \\ t &= \frac{\ln 2}{\lambda}, \end{aligned}$

so that the half-life of the substance is $(\ln 2)/{\lambda}$ . See this post for more exponential models.

For Fun. See this video for goated-era music. See this game for a game named after this idea.

Question 4. Solve the differential equation

$\displaystyle \frac{\mathrm d y}{\mathrm dx} + \frac yx = \frac{\sec(xy)}{x}.$

Hint: Use the substitution $u = xy$ .

(Click for Solution)

Solution. By the substitution $u = xy$ ,

$\displaystyle \frac{\mathrm du}{\mathrm dx} = y + x \frac{\mathrm dy}{\mathrm dx}.$

Multiplying the original equation by $x$ ,

$x\displaystyle \frac{\mathrm d y}{\mathrm dx} + y = \sec(xy).$

Therefore, the substitution $u = xy$ yields

$\displaystyle \frac{\mathrm du}{\mathrm dx} = \sec(u) = \frac{1}{\cos(u)}.$

By the method of separable variables,

$\begin{aligned} \frac{\mathrm du}{\mathrm dx} &= \frac{1}{\cos(u)} \\ \int \cos(u)\, \mathrm du &= \int 1\, \mathrm dx \\ \sin(u) &= x + C \\ \sin(xy) &= x + C.\end{aligned}$

—Joel Kindiak, 17 Apr 25, 1453H

April 20, 2025
The Mass-Spring System
Problem 1. An object with mass $m$ is loaded onto a mass-spring system. After loading the object that is at rest, the length of the spring is $s$ . Let $x = x(t)$ denote the displacement of the object at time $t$ . In Newtonian mechanics, we are given the following information:
- Hooke’s law tells us that the force that the spring acts on the object is $F_{\mathrm{S}}= -k(s+x)$ , where $k > 0$ is a spring constant of proportionality.
- Newton’s law of gravitation tells us that the gravitational force acting on the object is $F_{\mathrm{G}} = mg$ , where $g \approx 9.81$ is the gravitational constant of proportionality (at least on earth).
- The damping force is given by $F_{\mathrm{D}} = -\gamma x'$ , where $\gamma >0$ is a damping constant of proportionality.
Let the damping force be characterised by the constant $\gamma > 0$ and the force arising from the spring be characterised by $k > 0$ . Newton’s second law states that at any point in time,

$mx'' = F_{\mathrm{S}} + F_{\mathrm{G}} + F_{\mathrm{D}}.$

When setting up the equations, we have the boundary conditions $mg = ks$ .

Derive a homogeneous second-order differential equation in terms of $m, \gamma, k$ and $x$ and its derivatives. Hence, find an expression for $x$ in terms of $t$ in the case:
- $\gamma^2 - 4mk > 0$ ,
- $\gamma^2 - 4mk = 0$ ,
- $\gamma^2 - 4mk < 0$ .
What is the long-run behaviour of $x$ ? What would happen if there was no damping force (i.e. $\gamma = 0$ )?

(Click for Solution)

Solution. By Newton’s second law,

$mx'' = -k(s+x)+mg+(-\gamma x').$

Since $mg = ks$ ,

$mx'' = -kx-\gamma x' \quad \Rightarrow \quad mx'' + \gamma x' + kx = 0.$

Consider the auxiliary equation $mr^2 + \gamma r + k = 0$ , where we used $r$ to reduce ambiguity. By the quadratic equation, the roots $r_{\pm}$ are given by

$\displaystyle r_{\pm} = \frac{-\gamma \pm \sqrt{\gamma^2 - 4mk}}{2m}.$

In the case $\gamma^2 - 4mk > 0$ , the roots $r_{\pm}$ are real and distinct and negative. Thus,

$x(t) = C_1 e^{r_- t} + C_2 e^{r_+ t}.$

This is known as an overdamped system.

In the case $\gamma^2 - 4mk = 0$ , the roots $r_{\pm} = -\gamma/(2m)$ are real and repeated and negative. Thus,

$x(t) = (C_1 t + C_2)e^{rt}.$

This is known as a critically damped system.

In the case $\gamma^2 - 4mk < 0$ , the roots $r_{\pm} = p \pm iq$ are complex conjugates with $p = -\gamma/(2m) < 0$ . Thus,

$x(t) = e^{pt} (C_1 \sin(qt) + C_2 \cos(qt)).$

This is known as an underdamped system.

In all instances, the negative term $-r/(2m)$ induces all exponential terms to $\to 0$ , so that $x(t) \to 0$ as $t \to \infty$ . In the case $\gamma = 0$ , we only have the case $-4mk < 0$ , so that

$x(t) = C_1 \sin(qt) + C_2 \cos(qt).$

Indeed, the case $\gamma = 0$ corresponds to the differential equation $mx'' + kx = 0$ , which characterises a simple harmonic motion.

—Joel Kindiak, 7 Mar 25, 2048H
April 17, 2025

art, physics, science, technology
The Step Functions
Recall that we are interested in solving initial value problems of the form

$ay'' + by' + cy = Q(t),\quad y(0) = y_0, \quad y'(0) = y_0',$

where $y_0, y_0'$ are pre-determined constants. Initial value problems are basically differential equations, with some starting conditions (i.e. specified values of $y(0)$ and $y'(0)$ . It is called an initial value problem since we are specifying conditions at the initial moment, namely, $t = 0$ .

Many a time, the function $Q(x)$ takes the form of a step function.

Definition 1. Given distinct real numbers $a_0, a_1,\dots, a_n$ , a function $f$ is a step function if there exist $t_1 <\dots < t_n$ such that

$f(t) = \begin{cases} a_0, & t < t_1, \\ a_i, & t_i \leq t < t_{i+1},\quad 1 \leq i \leq {n-1}, \\ a_{n}, & t \geq t_n,\\ \end{cases}$

Example 1. The unit step function at $0$ is a step function, since

$U(t) = \begin{cases} 0, & t < 0, \\ 1, & t \geq 0. \end{cases}$

There are several generalisations of the idea described in Definition 1. Namely, in each interval, the function need not be constant (i.e. we can replace each $a_i$ with $f_i(t)$ and the function will still make sense). Nevertheless, we will stick with constants $a_i$ for simplicity.

The unit step function is itself of considerable interest. Firstly, we can shift it to obtain specific functions that match our problem. Next, we can truncate the unit step functions so that its integral is trivial to compute (it’s just the area of a rectangle).

Lemma 1. For any set $K$ , define the indicator function by

$\mathbb I_K(t) := \begin{cases} 1, & t \in K, \\ 0, & t \notin K. \end{cases}$

Then for any real $a < b$ ,
- $\mathbb I_{(-\infty, a)} = 1 - U_a$ ,
- $\mathbb I_{[a, b)} = U_a - U_b$ ,
- $\mathbb I_{[b,\infty)} =U_b$ .
Here, we denote $U_a(t) \equiv U(t-a)$ for brevity.

In particular, $\displaystyle \int_{-\infty}^\infty \mathbb I_{[a, b)}(t)\, \mathrm dt = b-a$ .

Proof. We prove $\mathbb I_{[a, b)} = U_a - U_b$ and leave the rest as exercises.

We need to prove that

$U(t- a) - U(t-b) = \begin{cases}0, & t < a, \\ 1, & a \leq t < b, \\ 0, & t \geq b.\end{cases}$

For $t < a$ , since $a < b$ , we have $t < b$ , so that

$U(t- a) - U(t-b) = 0 - 0 = 0.$

For $a \leq t < b$ ,

$U(t- a) - U(t-b) = 1 - 0 = 1.$

For $t \geq b$ , since $a < b$ , we have $t \geq a$ , so that

$U(t-a) - U(t-b) = 1 - 1 = 0.$

Therefore,

$U(t- a) - U(t-b) = \begin{cases}0, & t < a, \\ 1, & a \leq t < b, \\ 0, & t \geq b,\end{cases}$

as required.

Its Laplace transform is trivial to compute.

Theorem 1. For any real $a$ , $\mathcal L\{U_a(t)\} \equiv \mathcal L\{U(t-a)\} = e^{-as}/s$ . In particular,

$\displaystyle \mathcal L\{U(t)\} = \frac 1s = \mathcal L\{1\}.$

Proof. Plugging in the definition of the Laplace transform,

$\displaystyle \begin{aligned}\mathcal L\{U(t-a)\} &= \int_0^\infty U(t-a) e^{-st}\, \mathrm dt \\ &= \int_0^a U(t-a) e^{-st}\, \mathrm dt + \int_a^\infty U(t-a) e^{-st}\, \mathrm dt \\ &= \int_0^a 0 \cdot e^{-st}\, \mathrm dt + \int_a^\infty 1\cdot e^{-st}\, \mathrm dt \\ &= \int_a^\infty e^{-st}\, \mathrm dt = \left[\frac{e^{-st}}{-s}\right]_a^\infty \\ &= 0 - \frac{e^{-as}}{-s} = \frac{e^{-as}}{s}.\end{aligned}$

Finally, we can scale these functions accordingly to obtain functions that match our problem. When we combine these functions, we obtain a step function $f$ . The converse is true as well: every step function can be constructed by “combining” a bunch of unit step functions together.

Theorem 2. Let $f$ be a defined by

$f(t) = \begin{cases} a_0, & t < t_1, \\ a_i, & t_{i} \leq t < t_{i+1},\quad 1 \leq i \leq n-1, \\ a_n, & t \geq t_n,\\ \end{cases}$

For each $1 \leq i \leq n$ , define the $i$ -th jump by $\Delta a_i := a_i - a_{i-1}$ . Then

$\displaystyle f = a_0 + \Delta a_1 U_{t_i} + \cdots + \Delta a_n U_{t_n}.$

Proof. We will illustrate the case $n = 2$ for simplicity. The general case follows by careful bookkeeping.

Suppose

$f(t) = \begin{cases} a_0, & t < t_1, \\ a_1, & t_1 \leq t < t_2, \\ a_2, & t \geq t_2. \end{cases}$

Then adding the various components,

$\displaystyle \begin{aligned} f &= a_0(1 - U_{t_1}) + a_1 (U_{t_1} - U_{t_2}) + a_2U_{t_2} \\ &= a_0 - a_0 U_{t_1} + a_1 U_{t_1} - a_1 U_{t_2} + a_2 U_{t_2} \\ &= a_0 + (a_1 - a_0) U_{t_1} + (a_2 - a_1) U_{t_2} \\ &= a_0 + \Delta a_1 U_{t_1} + \Delta a_2 U_{t_2}. \end{aligned}$

These step functions can be used to define a powerful notion of integration known as Lebesgue integration. Of more immediate interest, its greatest utility in the context of differential equations is that its Laplace transform becomes trivial to compute.

Corollary 1. Let $f$ be a defined by

$f(t) = \begin{cases} a_0, & t < t_1, \\ a_i, & t_{i} \leq t < t_{i+1},\quad 1 \leq i \leq n-1, \\ a_n, & t \geq t_n,\\ \end{cases}$

Then

$\displaystyle \begin{aligned} \mathcal L\{f(t)\} &= a_0 \cdot \frac 1s + \sum_{i=1}^n \Delta a_i \cdot \frac{e^{-st_i}}{s}. \end{aligned}$

Finally, given a step function $f$ , how do we compute the following Laplace transform?

$\mathcal L\{f^2\} \equiv \mathcal L\{f^2(t)\} \equiv \mathcal L\{(f(t))^2\}$

We will consider $f = 5U_2 - 7U_3$ for simplicity. A direct computation yields

$f^2 = (5U_1 - 7U_2)^2 = 25U_1^2 - 70U_1U_2 + 49U_2^2.$

But how do we evaluate the products of unit step functions?

Theorem 3. For any real $a \leq b$ , $U_a U_b = U_b$ .

Proof. We observe that

$U_a \cdot U_b = \mathbb I_{[a,\infty)} \cdot \mathbb I_{[b,\infty)} = \mathbb I_{[a,\infty) \cap [b,\infty)} = \mathbb I_{[b,\infty)} = U_b.$

We can then compute the calculation.

Example 2. Let $f = 5U_2 - 7U_3$ be a step function. Then

$f^2 = 25U_2 - 21U_3$

so that $\displaystyle \mathcal L\{(f(t))^2\} = 25 \cdot \frac{e^{-2s}}{s} - 21 \cdot \frac{e^{-3s}}{s}$ .

To answer the question we started with: how do we solve initial value problems involving step functions? By taking Laplace transforms.

Example 3. Given the initial value problem

$\displaystyle y'' - 5 y' + 6y = U(t-2),\quad y(0)=y'(0) = 0$

and denoting $Y(s) \equiv \mathcal L\{y(t)\} \equiv \mathcal L\{y\}(s)$ , evaluate $Y(s)$ .

Solution. Taking Laplace transforms on all sides,

$\displaystyle \mathcal L\{y'' - 5 y' + 6y\} = \mathcal L\{ U(t-2) \}.$

By linearity,

$\displaystyle \mathcal L\{y''\} - 5 \mathcal L\{y'\} + 6 \mathcal L\{y\} = \mathcal L\{ U(t-2) \}.$

By the Laplace transform of the second derivative,

$\displaystyle \begin{aligned} \mathcal L\{y''\} &= s^2 Y(s) - s y(0) - y'(0) \\ &= s^2 Y(s) - 0 - 0 = s^2 Y(s). \end{aligned}$

By the Laplace transform of the first derivative,

$\displaystyle \begin{aligned} \mathcal L\{y'\} &= s Y(s) - y(0) \\ &= s Y(s) - 0 = s Y(s). \end{aligned}$

By notation, $\mathcal L\{y\} = Y(s)$ . By Theorem 1,

$\displaystyle \mathcal L \{ U(t-2) \} = \frac{e^{-2s}}s.$

Substituting all of these results,

$\displaystyle s^2 Y(s) - 5sY(s) + 6Y(s) = \frac {e^{-2s}}s.$

By basic algebruh,

$\displaystyle Y(s) = \frac{e^{-2s}}{s(s-2)(s-3)}.$

From this result, we still face another pressing problem. We will need to compute

$\displaystyle y(t) = \mathcal L^{-1}\{Y(s)\} = \mathcal L^{-1} \left\{ e^{-2s} \cdot \frac{1}{s(s-2)(s-3)} \right\}.$

How do we actually evaluate the expression on the right? This is where we will employ the help of inverse Laplace transforms and the shift theorems, in our next discussion.

—Joel Kindiak, 10 Feb 25, 1151H
April 15, 2025
More Laplace Transforms

Problem 1. Prove that for any real constant $a > 0$ ,

$\displaystyle \mathcal L\{ \sin (at) \} = \frac{a}{s^2+a^2},\quad \mathcal L\{ \cos (at) \} = \frac{s}{s^2+a^2}.$

(Click for Solution)

Solution. By Euler’s formula, $e^{i(at)} = \cos(at) + i \sin (at)$ . Taking Laplace transforms on both sides and applying linearity,

$\mathcal L\{ e^{i(at)} \} = \mathcal L\{ \cos (at) \} + i \cdot \mathcal L\{ \sin (at) \}.$

On the left-hand side,

$\displaystyle \begin{aligned} \mathcal L\{ e^{(ia)t} \} &= \frac{1}{s - ia} \\ &= \frac{1}{s-ia} \cdot \frac{s+ia}{s+ia} \\ &= \frac{s+ia}{s^2 + a^2} \\ &= \frac{s}{s^2 + a^2} + i \cdot \frac{a}{s^2 + a^2}.\end{aligned}$

Combining these results,

$\displaystyle \mathcal L\{ \cos (at) \} + i \cdot \mathcal L\{ \sin (at) \} = \frac{s}{s^2 + a^2} + i \cdot \frac{a}{s^2 + a^2}.$

Problem 2. Prove that for any $n$ , $\mathcal L\{t^n\} = n!/s^{n+1}$ .

(Click for Solution)

Solution. By induction, we can prove that for any reasonably-defined function $f$ ,

$\mathcal L\{ f^{(n)} (t) \} = s^n \mathcal L\{ f(t)\},$

if $f^{(k)}(0) = 0$ for $k = 0, 1, \dots, n-1$ .

Defining $f(t) = t^n$ , we have $f^{(k)}(0) = 0$ for $k = 0, 1, \dots, n-1$ and $f^{(n)}(t) = n!$ . Applying these results into the derivative result,

$\mathcal L\{ n! \} = s^n \mathcal L\{ t^n \}.$

Hence,

$\displaystyle \begin{aligned} \mathcal L\{ t^n \} &= \frac 1{s^n} \cdot \mathcal L\{ n! \} \\ &= \frac 1{s^n} \cdot n! \cdot \mathcal L\{ 1 \} \\ &= \frac 1{s^n} \cdot n! \cdot \frac 1s = \frac{n!}{s^{n+1}}. \end{aligned}$

Problem 3. Suppose there exists a function $\delta$ such that $U' = \delta$ . Prove that for any real constant $a > 0$ , $\mathcal L\{\delta(t - a)\} = e^{-as}$ .

(Click for Solution)

Solution. Assuming there exists such a function,

$\begin{aligned} \mathcal L\{\delta(t-a)\} &= \mathcal L\{U'(t-a) \cdot (t-a)'\} \\ &= \mathcal L\{(U(t-a))'\} \\ &= s \cdot \mathcal L\{U(t-a)\} - U(0 - a) \\ &= s \cdot \frac{e^{-as}}{s} - 0 = e^{-as}.\end{aligned}$

This notion, defined properly, is known as the Dirac delta distribution.

—Joel Kindiak, 13 Feb 25, 0152H

April 8, 2025
The Shift Theorems

In our previous write-up, we encountered a new problem in solving our initial value problem; how do we evaluate the expression below?

$\displaystyle \mathcal L^{-1} \left\{ e^{-2s} \cdot \frac{1}{s(s-2)(s-3)} \right\}$

Well, let’s first define $\mathcal L^{-1}$ properly.

Definition 1. For any function $F$ , we write $f(t) = \mathcal L^{-1}\{F(s)\}$ if there exists a function $f$ such that

$\mathcal L\{f(t)\} = F(s).$

Example 1. Based on previous discussions,

$\displaystyle \begin{aligned} \mathcal L^{-1}\{1/s\} &= 1, \\ \mathcal L^{-1}\{1/(s-a)\} &= e^{at}, \\ \mathcal L^{-1}\{e^{-as}/s\}&= U(t-a). \end{aligned}$

We notice that setting $a = 0$ in the last case, $1 = U(t)$ , which is false for $t < 0$ , since $1 \neq 0 = U(t)$ in that case. Nevertheless, as far as Laplace inverse transforms are concerned, they are the same function. The reason is that $1 = U(t)$ for $t \geq 0$ .

Hence, we will regard $f = g$ whenever $f(t) = g(t)$ for $t \geq 0$ . (The formal identification can be described using equivalence relations). These investigations, however, raises an interesting observation. Writing $f(t) = 1$ , we have

$\displaystyle F(s) \equiv \mathcal L\{f(t)\} = \frac 1s.$

Hence,

$\displaystyle \mathcal L\{e^{at} f(t)\} = \frac 1{s-a} = F(s-a).$

In fact, this result holds in general, and is known as the first of the shift theorems.

Theorem 1 (First Shift Theorem). For any real $a$ and function $f$ such that $\mathcal L\{f\}$ exists, $\mathcal L\{e^{at}f(t)\}$ exists and equals

$\displaystyle \mathcal L\{e^{at} f(t)\} = \frac 1{s-a} = F(s-a).$

Proof. For any $s$ ,

$\displaystyle \begin{aligned} \mathcal L\{e^{at} f(t)\} &= \int_0^\infty e^{at}f(t) \cdot e^{-st}\, \mathrm dt \\ &= \int_0^\infty f(t) \cdot e^{-(s-a)t}\, \mathrm dt \\ &= F(s-a). \end{aligned}$

This result is named the shift theorem because multiplying the exponential shifts the Laplace transform.

Can we reverse the question? That is, what would the expression $\mathcal L^{-1}\{e^{-as} F(s) \}$ evaluate to? Let’s explore this question. Suppose there exists $f,g$ such that $\mathcal L\{f\} = F$ and

$\mathcal L\{g(t)\} = e^{-as} F(s).$

When in doubt, return to the definitions—that would be first-principles thinking. Unpacking the definition of Laplace transforms yields

$\displaystyle \begin{aligned} \mathcal L\{g(t)\} &= e^{-as} \int_0^\infty f(t)e^{-st}\, \mathrm dt \\ &= \int_0^\infty f(t)e^{-s(a+t)}\, \mathrm dt \\ &= \int_a^\infty f(w-a)e^{-sw}\, \mathrm dw \\ &= \int_0^\infty f(w-a)U(w-a) e^{-sw}\, \mathrm dw \\ &= \int_0^\infty f(t-a)U(t-a) e^{-st}\, \mathrm dt \\ &= \mathcal L\{ f(t-a)U(t-a) \}.\end{aligned}$

Therefore, we can define $g(t) := f(t-a)U(t-a)$ , so that we obtain the second shift theorem below:

Theorem 2 (Second Shift Theorem). For any real $a$ and function $f$ such that $\mathcal L\{f\}$ exists, $\mathcal L\{e^{at}f(t)\}$ exists and equals

$\displaystyle \mathcal L^{-1} \{e^{-as} F(s)\} = f(t-a)U(t-a).$

Returning to our original problem, we want to find an expression for

$\displaystyle \mathcal L^{-1} \left\{ e^{-2s} \cdot \frac{1}{s(s-2)(s-3)} \right\}.$

This means that if we allow $\displaystyle F(s) = \frac{1}{s(s-2)(s-3)}$ and we can compute $f(t) = \mathcal L^{-1}\{F(s)\}$ , then by the second shift theorem, we have the following result.

Lemma 1. $\displaystyle \mathcal L^{-1} \left\{ e^{-2s} \cdot \frac{1}{s(s-2)(s-3)} \right\} = f(t-2)U(t-2).$

The only problem is: what is $f(t)$ ? Since $f(t) = \mathcal L^{-1}\{F(s)\}$ , we need to ask the question, how do we evaluate the following expression?

$\displaystyle \mathcal L^{-1} \left\{ \frac{1}{s(s-2)(s-3)} \right\}$

The denominator looks “separable”. Suppose we can, in fact, separate the fraction into a sum of partial fractions:

$\displaystyle \frac{1}{s(s-2)(s-3)} = \frac As + \frac B{s-2} + \frac C{s-3}.$

Then using the linearity of $\mathcal L$ , we can show that $\mathcal L^{-1}$ is linear too, so that

$\displaystyle \begin{aligned} &\mathcal L^{-1}\left\{ \frac{1}{s(s-2)(s-3)} \right\}\\ &= \mathcal L^{-1}\left\{ \frac As + \frac B{s-2} + \frac C{s-3} \right\} \\ &= A \cdot \mathcal L^{-1} \left\{ \frac 1s \right\} + B \cdot \mathcal L^{-1} \left\{ \frac 1{s-2} \right\} + C \cdot \mathcal L^{-1} \left\{ \frac 1{s-3} \right\}\\ &= A \cdot 1 + B \cdot e^{2t} + C \cdot e^{3t} \\ &= A + B \cdot e^{2t} + C \cdot e^{3t} .\end{aligned}$

Well, that is very convenient! All that remains is to compute $A, B, C$ , and we are done! Of course, there are several more loose ends to tie, such as the various denominators that can arise, but that would become a matter of bookkeeping and exploration of more formulas. Perhaps we will explore those other formulas in an exercise.

Back to computing $A, B, C$ . Beginning with the question at hand, we will clear denominators to obtain

$\displaystyle \begin{aligned} \frac{1}{s(s-2)(s-3)} &= \frac As + \frac B{s-2} + \frac C{s-3} \\ 1 &= A(s-2)(s-3) + Bs(s-3) + Cs(s-2). \end{aligned}$

Setting $s = 0$ yields $1 = A \cdot (-2) \cdot (-3) \Rightarrow A = 1/6$ .
Setting $s = 2$ yields $1 = B \cdot 2 \cdot (-1) \Rightarrow B = -1/2$ .
Setting $s = 3$ yields $1 = C \cdot 3 \cdot 1 \Rightarrow C = 1/3$ .

Back-substituting our results, we have the following expression for $f$ .

Lemma 2. $\displaystyle f(t) = \frac 16 - \frac 12 \cdot e^{2t} + \frac 13 \cdot e^{3t}.$

Thus, we can finally answer our main question as follows.

Example 2. Solve the initial value problem

$y'' - 5y' + 6y = U(t-2),\quad y(0) = 0, \quad y'(0) = 0.$

Solution. Taking Laplace transforms on all sides,

$\displaystyle \mathcal L\{y'' - 5 y' + 6y\} = \mathcal L\{ U(t-2) \}.$

By linearity,

$\displaystyle \mathcal L\{y''\} - 5 \mathcal L\{y'\} + 6 \mathcal L\{y\} = \mathcal L\{ U(t-2) \}.$

By the Laplace transform of the second derivative,

$\displaystyle \begin{aligned} \mathcal L\{y''\} &= s^2 Y(s) - s y(0) - y'(0) \\ &= s^2 Y(s) - 0 - 0 = s^2 Y(s). \end{aligned}$

By the Laplace transform of the first derivative,

$\displaystyle \begin{aligned} \mathcal L\{y'\} &= s Y(s) - y(0) \\ &= s Y(s) - 0 = s Y(s). \end{aligned}$

By notation, $\mathcal L\{y\} = Y(s)$ . By the Laplace transform of unit step functions,

$\displaystyle \mathcal L \{ U(t-2) \} = \frac{e^{-2s}}s.$

Substituting all of these results,

$\displaystyle s^2 Y(s) - 5sY(s) + 6Y(s) = \frac {e^{-2s}}s.$

By basic algebruh,

$\displaystyle Y(s) = \frac{e^{-2s}}{s(s-2)(s-3)}.$

Taking inverse Laplace transforms and applying Lemmas 1 and 2 respectively,

$\displaystyle \begin{aligned} y(t) &= \mathcal L^{-1}\{Y(s)\} \\ &= \mathcal L^{-1} \left\{ \frac{e^{-2s}}{s(s-2)(s-3)} \right\} \\ &= f(t-2) \cdot U(t-2) \\ &= \left(\frac 16 - \frac 12 \cdot e^{2(t-2)} + \frac 13 \cdot e^{3(t-2)} \right) \cdot U(t-2).\end{aligned}$

—Joel Kindiak, 10 Feb 25, 2207H

April 1, 2025
Who Cares about Laplace Transforms?

Consider the Heaviside step function, commonly denoted $U(\cdot)$ or $H(\cdot)$ , defined by

$U(t) \equiv \mathbb I_{[0,\infty)}(t) = \begin{cases} 1, & t \geq 0, \\ 0, & t < 0. \end{cases}$

How do we find the general solution for the differential equation below?

$\displaystyle \frac{\mathrm d^2 y}{\mathrm dt^2} - 5 \frac{\mathrm dy}{\mathrm dt} + 6y = U(t)$

We can write this in $\mathcal D$ -notation to obtain the equation

$(\mathcal D^2 - 5\mathcal D + 6)(y) = U(t),$

so perhaps we can find $y_{\mathrm C} = C_1 e^{2t} + C_2 e^{3t}$ and

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

Except that $y_{\mathrm P}$ isn’t immediately possible. Such conundrums motivate the Laplace transform as a powerful problem-solving tool.

Definition 1. Let $f$ be a function. For any $s$ , the Laplace transform of $f$ , denoted $\mathcal L\{f\}$ , is defined by

$\displaystyle \mathcal L\{f\}(s) := \int_0^{\infty} f(t)e^{-st}\, \mathrm dt \equiv \lim_{T \to \infty} \int_0^T f(t)e^{-st}\, \mathrm dt,$

whenever the right-hand side is well-defined. Defining $F = \mathcal L\{ f\}$ , we sometimes abuse notation and denote

$F = F(s) \equiv \mathcal L\{f(t)\} = \mathcal L\{ f \}.$

Example 1. For any $s > 0$ , $\mathcal L\{1\} = 1/s$ .

Proof. By the definition of Laplace transforms,

$\displaystyle \begin{aligned} \mathcal L\{1\} &= \int_0^{\infty} 1 \cdot e^{-st}\, \mathrm dt = \int_0^{\infty} e^{-st}\, \mathrm dt\\ &= \frac 1{-s}\left[ e^{-st} \right]_0^\infty = \frac{1}{-s}(0 - 1) = \frac 1s. \end{aligned}$

We will explore more commonly used Laplace transforms and connect them to commonly used solving techniques in future posts. Let’s first verify a familiar linearity property for the Laplace transforms that justifies the technique of “splitting”.

Theorem 1. Let $f,g$ be functions such that $\mathcal L\{f\}$ and $\mathcal L\{g\}$ exist, and $c$ be a constant. Then $\mathcal L\{f+g\} = \mathcal L\{f\} + \mathcal L\{g\}$ and $\mathcal L\{cf\} = c \mathcal L\{f\}$ both exist.

Proof. We prove the result for $\mathcal L\{f+g\} = \mathcal L\{f\} + \mathcal L\{g\}$ . By definition of the Laplace transform,

$\displaystyle \begin{aligned} \mathcal L\{f(t)\} &= \lim_{T \to \infty} \int_0^T f(t)e^{-st}\, \mathrm dt \end{aligned}$

By the linearity of taking integrals and taking limits,

$\displaystyle \begin{aligned} L\{(f+g)(t)\} &= \lim_{T \to \infty} \int_0^T (f(t) + g(t))e^{-st}\, \mathrm dt\\ &= \lim_{T \to \infty} \left(\int_0^T f(t)e^{-st}\, \mathrm dt + \int_0^T g(t)e^{-st}\, \mathrm dt \right) \\ &= \lim_{T \to \infty} \int_0^T f(t) e^{-st}\, \mathrm dt + \lim_{T \to \infty} \int_0^T g(t) e^{-st}\, \mathrm dt \\ &= \mathcal L\{f(t)\} + \mathcal L\{g(t)\}. \end{aligned}$

Coupled with linearity, we also need the Laplace transform of derivatives.

Theorem 2. Let $f$ be a differentiable function such that $F(s) = \mathcal L\{f(t)\}$ exists, $f'$ is integrable on any $[0, t]$ and $f(t)e^{-st} \to 0$ as $t \to \infty$ . Then $\mathcal L\{f'(t)\}$ exists and is given by

$\displaystyle \mathcal L\{f'(t)\} = s F(s) - f(0).$

Proof. Integrating by parts,

$\displaystyle \begin{aligned} \mathcal L\{f'(t)\} &= \int_0^\infty f'(t) e^{-st}\, \mathrm dt \\ &= [ \underbrace{f(t)}_{\mathrm I} \underbrace{e^{-st}}_{\mathrm S} ]_0^\infty - \int_0^\infty \underbrace{f(t)}_{\mathrm I} \cdot \underbrace{-se^{-st}}_{\mathrm D}\, \mathrm dt \\ &= (0 - f(0) \cdot 1) + s \int_0^\infty f(t)e^{-st}\, \mathrm dt\\&= - f(0) + s \mathcal L\{f(t)\} \\ &= sF(s) - f(0). \end{aligned}$

Corollary 1. Let $f$ be a twice-differentiable function such that $\mathcal L\{f(t)\}$ and $\mathcal L\{f'(t)\}$ exist, $f''$ is integrable on any $[0, t]$ , and $f(t)e^{-st} \to 0, f'(t)e^{-st} \to 0$ as $t \to \infty$ . Then $\mathcal L\{f''(t)\}$ exists and is given by

$\displaystyle \mathcal L\{f''(t)\} = s^2 F(s) - sf(0) - f'(0).$

Proof. Applying the result in Theorem 2 to $f'' = (f')'$ ,

$\begin{aligned} \mathcal L\{f''(t)\} &= s \cdot \mathcal L\{f'(t)\} - f'(0)\\ &= s \cdot (s \cdot \mathcal L\{f(t)\} - f(0)) - f'(0) \\ &= s^2 F(s) - sf(0) - f'(0). \end{aligned}$

These results give us the main power of the Laplace transform, which we will illustrate in the motivating example.

Example 2. Given the initial value problem

$\displaystyle y'' - 5 y' + 6y = U(t),\quad y(0)=y'(0) = 0$

and denoting $Y(s) \equiv \mathcal L\{y(t)\} \equiv \mathcal L\{y\}(s)$ , evaluate $Y(s)$ .

Solution. Taking Laplace transforms on all sides,

$\displaystyle \mathcal L\{y'' - 5 y' + 6y\} = \mathcal L\{ U(t) \}.$

By linearity,

$\displaystyle \mathcal L\{y''\} - 5 \mathcal L\{y'\} + 6 \mathcal L\{y\} = \mathcal L\{ U(t) \}.$

By Corollary 1,

$\displaystyle \begin{aligned} \mathcal L\{y''\} &= s^2 Y(s) - s y(0) - y'(0) \\ &= s^2 Y(s) - 0 - 0 = s^2 Y(s). \end{aligned}$

By Theorem 2,

$\displaystyle \begin{aligned} \mathcal L\{y'\} &= s Y(s) - y(0) \\ &= s Y(s) - 0 = s Y(s). \end{aligned}$

By notation, $\mathcal L\{y\} = Y(s)$ . By a result that we will discuss in the future, $\mathcal L \{ U(t) \} = 1/s$ . Substituting all of these results,

$\displaystyle s^2 Y(s) - 5sY(s) + 6Y(s) = \frac 1s.$

By basic algebruh,

$\displaystyle Y(s) = \frac{1}{s(s-2)(s-3)}.$

Now, we recall that $Y(s) = \mathcal L \{y(t)\}$ . How do we recover $y(t)$ ? Effectively, we would take the inverse Laplace transform

$y(t) = \mathcal L^{-1} \{Y(s)\},$

and employ several more tools at our disposal, like partial fractions, and more often than not, the shift theorems in the context of Laplace transforms. We will discuss these ideas more in a future post.

—Joel Kindiak, 5 Feb 25, 0041H

March 18, 2025
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
March 11, 2025
Differential Polynomials

We are officially asking harder questions. How do we solve differential equations that look like the following?

$\displaystyle a \frac{\mathrm d^2 y}{\mathrm dx^2} + b \frac{\mathrm d y}{\mathrm dx} + cy = Q(x)$

We have a straightforward result if $Q = 0$ . The fun arises when $Q \neq 0$ . The topics we handle now will assume $Q$ being differentiable, but eventually we will encounter bizarre types of $Q$ that required more exotic techniques.

How can we approach this systematically? What we will do is present a more compact version of what is known as the method of undetermined coefficients, but using a new tool called the inverse- $\mathcal D$ operator. What motivates such a definition? Well, recall that differentiation is a linear transformation in the following sense:

$\displaystyle \frac{\mathrm d}{\mathrm dx}(f + g) = \frac{\mathrm d}{\mathrm dx}(f) + \frac{\mathrm d}{\mathrm dx}(g), \quad \frac{\mathrm d}{\mathrm dx}(kf) = k \frac{\mathrm d}{\mathrm dx}(f).$

We can then interpret $\mathcal D$ as an operator whose input is a function $f$ and its output is another function $\mathcal D(f)$ . We define

$\displaystyle \mathcal D(f):= \frac{\mathrm d}{\mathrm dx}(f)$

so that the linearity property of $\displaystyle \mathcal D \equiv \frac{\mathrm d}{\mathrm dx}$ can be written as

$\mathcal D(f + g) = \mathcal D(f) + \mathcal D(g), \quad \mathcal D(kf) = k \mathcal D(f).$

Definition 1. For any $n \in \mathbb N_0$ ,

$\displaystyle \mathcal D^n(f) = \begin{cases} f, & n = 0, \\ \mathcal D(\mathcal D^{n-1}(f)), & n > 0. \end{cases}$

Example 1. For any $k \in \mathbb C \supseteq \mathbb R$ , the following hold:

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

Furthermore, for any $n \in \mathbb N_0$ , $\mathcal D^n(e^{kx}) = k^n e^{kx}$ .

Using $\mathcal D$ -notation, the original differential equation can be simplified as follows

$\begin{aligned} Q(x) &= a \frac{\mathrm d^2 y}{\mathrm dx^2} + b \frac{\mathrm d y}{\mathrm dx} + cy \\ &= a \mathcal D^2 (y) + b \mathcal D (y) + cy \\ &= (a \mathcal D^2 + b \mathcal D + c)y, \end{aligned}$

where the last line follows from adding operators according to Definition 2 below.

Definition 2. Let $p(x) = a_0 + a_1x + \cdots + a_nx^n$ , $a_n \neq 0$ be a polynomial with real coefficients $a_i$ . Define the differential polynomial by

$p(\mathcal D) y \equiv (p(\mathcal D))y := a_0 + a_1 \mathcal D(y) + \cdots + a_n \mathcal D^n(y).$

By the definition of adding functions,

$p(\mathcal D) \equiv a_0 + a_1 \mathcal D + \cdots + a_n \mathcal D^n$

Theorem 1. For any polynomial $p(x)$ , $p(\mathcal D)$ is a linear transformation in the following sense:

$p(\mathcal D)(f+g) = p(\mathcal D)(f) + p(\mathcal D)(g),\quad p(\mathcal D)(kf) = k p(\mathcal D)(f).$

Proof. The properties hold for operators of the form ${\mathcal D}^k$ . Let $\mathcal D_1, \mathcal D_2$ denote operators that satisfy the linearity properties. Checking that scaling and adding such operators that preserve linearity then yields the result:

$\begin{aligned} (c \mathcal D_1)(f+g) &= c(\mathcal D_1(f)+\mathcal D_1(g)) \\ &= (c \mathcal D_1) (f) + (c \mathcal D_1) (g),\\ (\mathcal D_1 + \mathcal D_2)(f+g) &= (\mathcal D_1)(f+g) + (\mathcal D_2)(f+g) \\ &= \mathcal D_1(f) + \mathcal D_1(g) + \mathcal D_2(f) + \mathcal D_2(g) \\ &= \mathcal D_1(f) + \mathcal D_2(f) + \mathcal D_1(g) + \mathcal D_2(g) \\ &= (\mathcal D_1 + \mathcal D_2)(f) + (\mathcal D_1 + \mathcal D_2)(g), \end{aligned}$

since $\displaystyle p(\mathcal D) = a_0 + a_1\mathcal D + \cdots + a_n\mathcal D^n$ .

Remark 1. These properties can be generalised and studied in more detail in the undergraduate course linear algebra.

Thus, we are interested to solve differential equations of the form

$p(\mathcal D)(y) = Q(x),$

where $p(x) = ax^2 + bx + c$ for real coefficients $a,b,c$ . We understand the situation when $Q = 0$ , and call such solutions the complementary functions, denoted $y_{\mathrm C}$ so that

$p(\mathcal D)(y_{\mathrm C}) = 0.$

It turns out that if we can find just one function $y_{\mathrm P}$ such that $p(\mathcal D)(y_{\mathrm P}) = Q(x)$ , then we can find all functions $y$ such that $p(\mathcal D)(y) = Q(x)$ . This approach isn’t unique to differential polynomials, but applies to any linear transformation.

Nevertheless, we will state this result in the context of differential polynomials, since any broader result belongs to linear algebra, rather than this topic.

Theorem 2. Suppose we have found the general solution to the equation $p(\mathcal D)(y_{\mathrm C}) = 0$ , called the complementary function $y_{\mathrm C}$ , and one particular integral $y_{\mathrm P}$ such that $p(\mathcal D)(y_{\mathrm P}) = Q(x)$ . Then the general solution to the equation $p(\mathcal D)(y) = Q(x)$ is given by $y = y_{\mathrm P} + y_{\mathrm C}$ .

Proof. Suppose $p(\mathcal D)(y) = Q(x)$ . Since $p(\mathcal D)(y_{\mathrm P}) = Q(x)$ and $p(\mathcal D)$ is a linear transformation,

$\displaystyle p(\mathcal D)(y-y_{\mathrm P}) = p(\mathcal D)(y) - p(\mathcal D)(y_{\mathrm P}) = Q(x) - Q(x) = 0.$

Since $p(\mathcal D)(u) = 0 \Rightarrow u = y_{\mathrm C}$ , we have

$y - y_{\mathrm P} = y_{\mathrm C} \quad \Rightarrow \quad y = y_{\mathrm P} + y_{\mathrm C}.$

Example 2. Find the general solution to the differential equation $\mathcal D(y) = x^2$ .

Solution. Given, $\mathcal D(y) = 0$ , we want to find all $y_{\mathrm C}$ such that $\displaystyle \frac{\mathrm d}{\mathrm dx}(y_{\mathrm C}) = 0$ . This means $y_{\mathrm C} = C$ for some real constant $C$ .

Recalling integration and differentiation as reverses of each other,

$\displaystyle \mathcal D\left( \frac 13 y^3 \right) = \frac 13 \cdot 3y^2 = y^2 \neq K\cdot C.$

Thus, we have found one $y_{\mathrm P} = y^3/3$ . Note that this choice is by no means unique.

The general solution, therefore, is $y = y_{\mathrm P} + y_{\mathrm C} = \frac 13 y^3 + C$ , which is what we would have obtained when solving

$\displaystyle \frac{\mathrm d}{\mathrm dx}(y) = \mathcal D(y) = x^2 \quad \iff \quad y = \int x^2\, \mathrm dx$

in the traditional manner.

Example 3. Find the general solution to the differential equation

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

Solution. We first solve the equation $(\mathcal D^2 - 5 \mathcal D + 6)(y) = 0$ . The characteristic equation $m^2 - 5m + 6 = 0$ yields the real and distinct roots $m = 2, 3$ . Hence,

$y_{\mathrm C} = C_1 e^{2x} + C_2 e^{3x}.$

For $y_{\mathrm P}$ , we just need one example that isn’t a combination of $e^{2x}$ or $e^{3x}$ . Our educated guess (the technical term in the business is called an Ansatz) here is then $y_{\mathrm P} = Ae^{kx}$ , where we will determine the (presently-undetermined) coefficients. We would like

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

If we can find $A$ and $k$ such that $y_{\mathrm P} = Ae^{kx}$ , the left-hand side simplifies to

$\begin{aligned} (\mathcal D^2 - 5\mathcal D + 6)(y_{\mathrm P}) &= \mathcal D^2 (y_{\mathrm P}) - 5 \cdot \mathcal D (y_{\mathrm P}) + 6 \cdot y_{\mathrm P} \\ &= A \cdot k^2 e^{kx} - 5 \cdot Ake^{kx} + 6 \cdot Ae^{kx} \\ &= A(k^2 - 5k + 6)e^{kx}.\end{aligned}$

This yields the equation

$A(k^2 - 5k + 6)e^{kx} = e^x.$

Hence, what would a good choice of $A$ and $k$ be? Well, we want the exponential terms to match, so $k = 1$ . Furthermore, we want the coefficients of the terms on both sides to match, so we require

$A(k^2 - 5k + 6) = 1 \quad \Rightarrow \quad A = 1/2.$

Hence, $y_{\mathrm P} = \frac 12 e^x$ will be a good choice. Combining our results, the general solution will then be

$\displaystyle y =y_{\mathrm P} + y_{\mathrm C} = \frac 12 e^x + C_1 e^{2x} + C_2 e^{3x}.$

Example 4. Find the general solution to the differential equation

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

Solution. By Example 3, we still have the complementary function

$y_{\mathrm C} = C_1 e^{2x} + C_2 e^{3x}.$

If we used the same Ansatz $y_{\mathrm P} = Ae^{kx}$ , then the default equation

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

simplifies to

$A(k^2 - 5k + 6)e^{kx} = e^{2x}.$

The same line of reasoning will yield $k = 2$ , but the coefficients won’t match, since $A(k^2 - 5k + 6) = 0 \neq 1$ . Clearly, the Ansatz $y_{\mathrm P} = Ae^{kx}$ won’t work here.

As if it’s some magical trick spawned from on-high, let’s try the Ansatz $y_{\mathrm P} = Axe^{kx}$ . Differentiating by using the product rule,

$\begin{aligned} \mathcal D(y_{\mathrm P}) &= Ae^{kx} + Akxe^{kx} = Ae^{kx} + k \cdot y_{\mathrm P}, \\ \mathcal D^2(y_{\mathrm P}) &= A\cdot ke^{kx} + k \cdot \mathcal D(y_{\mathrm P}) = 2Ake^{kx} + k^2 y_{\mathrm P}. \end{aligned}$

So the left-hand side of the default equation simplifies to

$\begin{aligned} e^{2x} &= (\mathcal D^2 - 5\mathcal D + 6)(y_{\mathrm P}) \\ &= (2Ake^{kx} + k^2 y_{\mathrm P}) - 5(Ae^{kx} + k \cdot y_{\mathrm P}) + 6 \cdot y_{\mathrm P} \\ &= (2k - 5)Ae^{kx} + (k^2 - 5k + 6) y_{\mathrm P}. \end{aligned}$

Now things look much nicer. Setting $k = 2$ will still yield $k^2 - 5k + 6 = 0$ . But now there is a nonzero term that will solve the problem, since

$e^{2x} = (2 \cdot 2 - 5)Ae^{2x} = (-A)e^{2x}.$

The appropriate choice for $A$ is then $-A = 1 \iff A = -1$ . Hence, we will choose $y_{\mathrm P} = -xe^{2x}$ to satisfy our requirements, yielding the general solution

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

Theorem 2 therefore provides us an underlying strategy for solving second-order differential equations with constant coefficients. In fact, we demonstrated the simplest form of the method of undetermined coefficients—by determining the coefficients in Examples 3 and 4. However, we would like to accomplish this goal in a more systematic manner. The working especially in Example 4 is exceedingly tedious. The challenge here arises from selecting a $y_{\mathrm P}$ that works; can we streamline this process?

The trick is to identify and tabulate the common functions for $y_{\mathrm P}$ and ensure that they satisfy our requirements. That is where inverse- $\mathcal D$ operators come in clutch.

—Joel Kindiak, 30 Jan 25, 1915H

February 25, 2025
Two-Dimensional Differential Equations
A first-order differential equation involves the first derivative $\displaystyle \frac{\mathrm dy}{\mathrm dx}$ . A second-order differential equation involves the second derivative $\displaystyle \frac{\mathrm d^2 y}{\mathrm dx^2}$ and even possibly the first.

Example 1. Given $\alpha \neq \beta$ , find the general solution to the equation

$\displaystyle \frac{\mathrm d^2 y}{\mathrm dx^2} - (\alpha + \beta) \frac{\mathrm d y}{\mathrm dx} + (\alpha \beta) y = 0.$

Solution. We will leave the solution as a guided exercise. Use the substitution $\displaystyle u = \frac{\mathrm dy}{\mathrm dx} - \beta y$ to obtain the differential equation

$\displaystyle \frac{\mathrm du}{\mathrm dx} - \alpha u = 0.$

By the method of separable variables, $u = Ae^{\alpha x}$ , so that

$\displaystyle \frac{\mathrm dy}{\mathrm dx} - \beta y = Ae^{\alpha x}.$

By the method of integrating factors,

$y = C_1 e^{\alpha x} + C_2 e^{\beta x}$ .

This yields the first result for our discussion on second-order differential equations.

Theorem 1. Given real constants $a, b, c$ , the general solution to the differential equation

$\displaystyle a \frac{\mathrm d^2 y}{\mathrm dx^2} + b \frac{\mathrm dy}{\mathrm dx} + c y = 0$

is determined by the roots of its characteristic equation or auxiliary equation $am^2 + bm + c = 0$ :
- $y = C_1 e^{\alpha x} + C_2 e^{\beta x}$ if the roots $\alpha \neq \beta$ are real and distinct.
- $y = (C_1x + C_2) e^{\alpha x}$ if the roots $\alpha = \beta$ are real and repeated.
- $y = e^{px}(C_1 \cos qx + C_2 \sin qx)$ if the roots $p \pm iq$ are complex conjugates.
Proof. We have proven the first case in Example 1. For the third case, we apply the first case to the observation

$e^{(p \pm iq)x} = e^{px}e^{\pm i(qx)} = e^{px} (\cos qx \pm i \sin qx),$

where the second equality follows from Euler’s formula. The solutions are then given by

$\displaystyle \begin{aligned}y &= C_1' e^{(p+iq)x} + C_2' e^{(p-iq)x} \\ &= e^{px} ((C_1' + C_2') \cos qx + i(C_1' - C_2') \sin qx) \\ &= e^{px} (C_1 \cos qx +C_2 \sin qx),\end{aligned}$

where $C_1 := C_1'+C_2'$ and $C_2 := i(C_1'-C_2')$ are constants.

For the second case, the derivation of the first case leads to the differential equation

$\displaystyle \frac{\mathrm dy}{\mathrm dx} - \alpha y = Ae^{\alpha x}.$

By the method of integrating factors, we have an integrating factor $\mu(x) = e^{\int -\alpha\,\mathrm dx} = e^{-\alpha x}$ so that the general solution is given by

$\displaystyle \begin{aligned} y \mu(x) &= \int \mu(x) \cdot Ae^{\alpha x}\, \mathrm dx\\ y e^{-\alpha x} &= \int e^{-\alpha x} \cdot Ae^{\alpha x}\, \mathrm dx \\ &= \int A \, \mathrm dx \\ &= Ax + C \\ y &= (Ax + C)e^{\alpha x} \\ &= (C_1 x + C_2) e^{\alpha x}, \end{aligned}$

where $C_1:= A, C_2 := C$ are constants.

For naming purposes:
- Since the left-hand side is of the form $\displaystyle (*) \frac{\mathrm d^2 y}{\mathrm dx^2} + (*) \frac{\mathrm d y}{\mathrm dx} + (*) y$ , the ODE is called linear.
- Since the highest derivative is $\displaystyle \frac{\mathrm d^2 y}{\mathrm dx^2}$ , the ODE is of second-order.
- Since the coefficients in each $(*)$ are constants $a, b, c$ , the ODE is said to have constant coefficients.
- Since the right-hand side is $0$ , the ODE is said to be homogeneous.
Putting it all together, differential equations of the form

$\displaystyle a \frac{\mathrm d^2 y}{\mathrm dx^2} + b \frac{\mathrm dy}{\mathrm dx} + cy = 0$

are called homogeneous second-order linear ODEs with constant coefficients (what a mouthful…).

Example 2. Given $\omega > 0$ , an object is said to obey simple harmonic motion if its acceleration $\displaystyle \frac{\mathrm d^2 x}{\mathrm dt^2}$ is directly proportional to its displacement $x$ with proportionality constant $-\omega^2$ .

Given that the object obeying simple harmonic motion has initial displacement $x_0$ and initial velocity $v_0$ , obtain an expression for $x$ in terms of $\omega$ , $t$ , $x_0$ , $v_0$ .

Solution. Since the object obeys simple harmonic motion,

$\displaystyle \frac{\mathrm d^2 x}{\mathrm d t^2} = -\omega^2 x \quad \Rightarrow \quad \frac{\mathrm d^2 x}{\mathrm d t^2} + \omega^2 x = 0.$

The characteristic equation is given by $m^2 + \omega^2 = 0$ and its roots are $m = 0\pm i\omega$ . Thus, the general solution of the differential equation is given by

$\displaystyle x = C_1 \cos \omega t + C_2 \sin \omega t.$

Since $x(0) = x_0$ yields $C_1 = x_0$ and $x'(0) = v_0$ yields $C_2 = v_0/\omega$ , we have

$\displaystyle x = x_0 \cos \omega t + \frac{v_0}{\omega} \sin \omega t.$

Things start getting rather challenging when the right-hand side of our differential equation is not $0$ ; how do we solve the following differential equation?

$\displaystyle a \frac{\mathrm d^2 y}{\mathrm dx^2} + b \frac{\mathrm dy}{\mathrm dx} + c y = f(x)$

In fact, we haven’t even considered the cases where $a,b,c$ are replaced by functions themselves. This is itself a nontrivial area of study, which we may explore if time and space permits.

Nevertheless, working with constant $a$ , $b$ , $c$ , we have several approaches to solve the corresponding differential equation. We will explore both approaches in due time; the Laplace approach in its traditional presentation, and the undetermined coefficients in a rather compact presentation involving inverse- $\mathcal D$ operators.

These approaches by no means solve every differential equation out there, but it does suffice to analytically solve many equations that are in use in the STEM fields. Most differential equations are solved numerically anyway, and then we invoke the pure mathematicians’ insanity to verify that these solutions genuinely exist. But that’s a discussion for another day.

—Joel Kindiak, 28 Jan 25, 1616H
February 18, 2025
Differential Equations Recap (Polytechnic)

First-Order Differential Equations

Problem 1. Solve the separable differential equation $\displaystyle \frac{\mathrm dy}{\mathrm dx} = x^2 y$ .

(Click for Solution)

Solution. By the method of separable variables,

$\begin{aligned} \int \frac 1y\, \mathrm dy &= \int x^2\, \mathrm dx,\\ \ln |y| &= \frac{x^3}{3} + C \\ |y| &= e^{x^3/3 + C} \\ y &= \pm e^C \cdot e^{x^3 / 3} \\ y &= Ae^{x^3/3},\end{aligned}$

where we consolidated the arbitrary constant $A = \pm e^C$ .

Problem 2. Solve the linear differential equation $\displaystyle \frac{\mathrm dy}{\mathrm dx} + 2y = e^{3x}$ .

(Click for Solution)

Solution. By the method of integrating factors, we identify $P(x) = 2$ and $Q(x) = e^{3x}$ . The integrating factor $I(x)$ is computed via

$I(x) = e^{\int P(x)\, \mathrm dx} = e^{\int 3\, \mathrm dx} = e^{3x}.$

(In our notes, we denoted $\mu(x) \equiv I(x)$ . This solution uses the notation that is used in the school that I teach.) The general solution is then given by

$\begin{aligned} y I(x) &= \int I(x)Q(x)\, \mathrm dx \\ y \cdot e^{2x} &= \int e^{3x} \cdot e^{2x}\, \mathrm dx\\ &= \int e^{5x}\, \mathrm dx\\ &= \frac 15 e^{5x} + C \end{aligned}$

Second-Order Differential Equations

Problem 3. Solve the differential equation $y'' - 5y' + 6y = 0$ .

(Click for Solution)

Solution. The characteristic equation

$m^2 - 5m + 6 = 0$

has solutions $m = 2, 3$ , which are real and distinct. Therefore, the general solution is given by

$y = C_1 e^{2x} + C_2 e^{3x}.$

Problem 4. Evaluate $\displaystyle \frac 1{\mathcal D^2 + 1} ( 2 + e^{2x} + \cos(x) + \sin(2x) )$ .

(Click for Solution)

Solution. By the linearity of inverse- $\mathcal D$ operators,

$\begin{aligned} &\frac 1{\mathcal D^2 + 1}\left( + e^{2x} + \cos(x) + \sin(2x) \right) \\ &= \frac 1{\mathcal D^2 + 1}(2) + \frac 1{\mathcal D^2 + 1} (e^{2x}) + \frac 1{\mathcal D^2 + 1}(\cos(x)) + \frac 1{\mathcal D^2 + 1}(\sin(2x)) \\ &= \frac 1{0^2 + 1} \cdot 2 + \frac 1{2^2 + 1} \cdot e^{2x} + \frac{x \sin(x)}{2 \cdot 1} + \frac 1{-2^2 + 1} \cdot \sin(2x) \\ &= 2 + \frac 1{5} e^{2x} + \frac{1}{2}x \sin(x) - \frac 1{3} \sin(2x). \end{aligned}$

Problem 5. Evaluate $\displaystyle \frac 1{\mathcal D^2 - 4\mathcal D - 5}(e^{2x} \sin(x))$ .

(Click for Solution)

Solution. Applying the shift property for inverse- $\mathcal D$ operators,

$\begin{aligned}\frac 1{\mathcal D^2 - 4\mathcal D - 5}(e^{2x} \sin(x)) &= e^{2x} \cdot \frac 1{(\mathcal D + 2)^2 - 4(\mathcal D + 2) - 5}(\sin(x)) \\ &= e^{2x} \cdot \frac 1{(\mathcal D^2 + 4 \mathcal D + 4) - (4\mathcal D + 8) - 5}(\sin(x)) \\ &= e^{2x} \cdot \frac 1{\mathcal D^2 - 9}(\sin(x)) \\ &= e^{2x} \cdot \frac 1{-1^2 - 9} \cdot \sin(x) \\ &= -\frac 1{10} e^{2x} \sin(x). \end{aligned}$

Problem 6. Solve the differential equation $\displaystyle \frac{\mathrm d^2 y}{\mathrm d x^2} - 4 \frac{\mathrm d y}{\mathrm d x} - 5y = 10 e^{2x} \sin(x)$ .

(Click for Solution)

Solution. To obtain the complementary function $y_{\mathrm C}$ , we solve the equation

$\displaystyle \frac{\mathrm d^2 y}{\mathrm d x^2} - 4 \frac{\mathrm d y}{\mathrm d x} - 5y = 0.$

The characteristic equation

$m^2 - 4m - 5 = 0$

has solutions $m = -1, 5$ , which are real and distinct. Therefore, the general solution is given by

$y = C_1 e^{-x} + C_2 e^{5x}.$

For the particular integral, we re-write the problem using $\mathcal D$ -notation

$(\mathcal D^2 - 4\mathcal D - 5)y_{\mathrm P} = 10 e^{2x} \sin(x)$

and use Problem 5 to obtain

$\begin{aligned} y_{\mathrm P} &= \frac 1{\mathcal D^2 - 4\mathcal D - 5} (10 e^{2x} \sin(x)) \\ &= 10 \cdot \frac 1{\mathcal D^2 - 4\mathcal D - 5} (e^{2x} \sin(x)) \\ &= 10 \cdot -\frac 1{10} e^{2x} \sin(x) \\ &= -e^{2x} \sin(x). \end{aligned}$

Consolidating, the general solution is given by

$y = y_{\mathrm P} + y_{\mathrm C} = -e^{2x} \sin(x) + C_1 e^{-x} + C_2 e^{5x}.$

Laplace Transforms

Problem 7. Evaluate $\mathcal L\{3 + 2e^{3t} + t^3 + \sin (2t)\}$ .

(Click for Solution)

Solution. By the linearity of taking Laplace transforms,

$\begin{aligned} &\mathcal L\{3 + 2e^{3t} + t^3 + \sin (2t)\} \\ &= 3 \cdot \mathcal L\{ 1 \} + 2 \cdot \mathcal L\{ e^{3t} \} + \mathcal L \{ t^3 \} + \mathcal L \{ \sin(2t) \} \\ &= 3 \cdot \frac 1s + 2 \cdot \frac 1{s-3} + \frac{3!}{s^{3+1}} + \frac 2{s^2+2^2} \\ &= \frac 3s + \frac 2{ s-3 } + \frac{ 6 }{ s^4 } + \frac 2{ s^2+4 }. \end{aligned}$

Problem 8. Evaluate $\displaystyle \mathcal L^{-1}\left\{ \frac 2{s^2} + \frac 2{s-1} + \frac 4{s^2+4} + \frac{5s}{s^2 + 16} \right\}$ .

(Click for Solution)

Solution. By the linearity of taking inverse Laplace transforms,

$\begin{aligned} & \mathcal L^{-1}\left\{ \frac 2{s^2} + \frac 2{s-1} + \frac 4{s^2+4} + \frac{5s}{s^2 + 16} \right\} \\ &= 2 \cdot \mathcal L^{-1}\left\{ \frac {1!}{s^{1+1}} \right\} + 2 \cdot \mathcal L^{-1}\left\{ \frac 1{s-1} \right\} \\ &\phantom{==} + 2 \cdot \mathcal L^{-1}\left\{ \frac 2{s^2+2^2} \right\} + 5 \cdot \mathcal L^{-1}\left\{ \frac s{s^2+4^2} \right\} \\ &= 2 t + 2 e^t + 2 \sin(2t) + 5 \cos(4t). \end{aligned}$

Problem 9. The function $f$ is defined by

$f(t) = \begin{cases} 0, & t < 2, \\ 3, & 2 \leq t < 5, \\ 4, & 5 \leq t < 7, \\ 2, & t \geq 7.\end{cases}$

Evaluate $\mathcal L\{e^{2t}f(t)\}$ .

(Click for Solution)

Solution. By the shift theorems,

$\mathcal L\{e^{2t} f(t)\} = F(s - 2),$

where $\mathcal L\{f(t)\} = F(s)$ . To evaluate $f(t)$ , we use the jump technique to obtain

$\begin{aligned} f(t) &= 0 + (3-0) \cdot U(t-2) + (4-3) \cdot U(t-5) + (2-4) \cdot U(t-7) \\ &= 3 U(t-2) + U(t-5) - 2 U(t-7). \end{aligned}$

By the linearity of taking Laplace transforms,

$\begin{aligned} F(s) = \mathcal L \{f(t) \} &= \mathcal L \{ 3 U(t-2) + U(t-5) - 2 U(t-7) \} \\ &= 3 \cdot \mathcal L \{ U(t-2) \} + \mathcal L \{ U(t-5) \} - 2 \cdot \mathcal L \{U(t-7) \} \\ &= 3 \cdot \frac{e^{-2s}}{s} + \frac{e^{-5s}}{s} - 2 \cdot \frac{e^{-7s}}{s} \\ &= \frac{3e^{-2s} + e^{-5s} - 2e^{-7s}}{s}. \end{aligned}$

Therefore,

$\displaystyle \mathcal L\{e^{2t} f(t)\} = F(s - 2) = \frac{3e^{-2(s-2)} + e^{-5(s-2)} - 2e^{-7(s-2)}}{s-2}.$

Problem 10. Solve the initial value problem

$y'' + y = \delta(t-2),\quad y(0)=y'(0) = 0.$

(Click for Solution)

Solution. Taking Laplace transforms on both sides and applying linearity,

$\begin{aligned} \mathcal L\{ y'' \} + \mathcal L \{ y \} &= \mathcal L \{ \delta(t-2) \} \end{aligned}$

To evaluate each expression, $\mathcal L\{y\} = Y(s)$ , $\mathcal L \{ \delta(t-2) \} = e^{-2s}$ , and

$\mathcal L\{ y'' \} = s^2 Y(s) - sy(0) - y'(0) = s^2 Y(s).$

Hence,

$\begin{aligned} \mathcal L\{ y'' \} + \mathcal L \{ y \} &= \mathcal L \{ \delta(t-2) \} \\ s^2 Y(s) + Y(s) &= e^{-2s} \\ (s^2 + 1) Y(s) &= e^{-2s} \\ Y(s) &= e^{-2s} \cdot \frac{1}{s^2+1}. \end{aligned}$

To obtain $y(t)$ , we take Laplace transforms to obtain

$\displaystyle y(t) = \mathcal L^{-1} \{ Y(s) \} = \mathcal L^{-1} \left\{ e^{-2s} \cdot \frac{1}{s^2+1} \right\}.$

To evaluate the right-hand side, let $F(s) = 1/(s^2+1)$ . By the shift theorems,

$\displaystyle y(t) = \mathcal L^{-1} \left\{ e^{-2s} \cdot \frac{1}{s^2+1} \right\} = f(t-2) U(t-2).$

This result requires us to obtain $f(t)$ , which can be done using inverse Laplace transforms:

$\displaystyle f(t) = \mathcal L^{-1} \{F(s)\} = \mathcal L^{-1} \left\{ \frac 1{s^2+1} \right\} = \sin(t).$

Hence, consolidating the results,

$\displaystyle y(t) = f(t-2)U(t-2) = \sin(t-2) U(t-2).$

—Joel Kindiak, 17 Feb 25, 0939H

February 17, 2025