Section Writing the Equation of a Line in Point-Slope Form
When asked for the general form for the equation of a line, most people give the familiar
\begin{equation*}
y = mx + b
\end{equation*}
βthe slope-intercept form. This is a great form, but there is another equally useful one, easy to remember if we recall the slope formula. Given two points \((x_1, y_1)\) and \((x, y)\text{,}\) the slope of the line between them is:
\begin{equation*}
m = \frac{y - y_1}{x - x_1}
\end{equation*}
Rearranging gives:
\begin{equation*}
m(x - x_1) = y - y_1
\end{equation*}
or, more commonly written:
\begin{equation*}
y - y_1 = m(x - x_1).
\end{equation*}
Sometimes itβs useful to go one step further and solve for \(y\) by adding \(y_1\) to both sides:
\begin{equation*}
y = m(x - x_1) + y_1.
\end{equation*}
π Example 305. Writing a Line from Two Points.
Given the points \((3, 7)\) and \((-1, 2)\text{,}\) (i) write the equation of the line through the points in point-slope form, and then (ii) solve for the dependent variable \(y\text{.}\)Solution.
First, compute the slope:
\begin{align*}
m \amp = \frac{2 - 7}{-1 - 3} = \frac{5}{4}
\end{align*}
Now write the point-slope form:
\begin{align*}
y - y_1 \amp = m(x - x_1)
\end{align*}
\begin{align*}
y - 7 \amp = \frac{5}{4}(x - 3) \quad \text{OR} \quad y - 2 = \frac{5}{4}(x + 1)
\end{align*}
(ii) Solve for
\(y\text{:}\)
\begin{align*}
y \amp = \frac{5}{4}(x - 3) + 7 \quad \text{OR} \quad y = \frac{5}{4}(x + 1) + 2
\end{align*}
π Example 306. Practice Writing Equations of Lines.
Use the given points to: (i) write the equation, in point-slope form, of the line through the points, and then (ii) solve for the dependent variable \(y\text{.}\)You have attempted
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Section Function Notation
Function notation is a cornerstone of mathematics and science because it allows us to describe relationships between variables in a compact and precise way. While seemingly simple, its power lies in its flexibility and depth. If youβve studied calculus, youβre familiar with the concept of a function, but reviewing its nuances can help unlock a deeper understanding of differential equations and mathematical modeling.
A function, in essence, is a rule that assigns to each input value exactly one output value. For example, the function
\(f(x) = x^2\) takes any input
\(x\text{,}\) squares it, and returns the result. The notation
\(f(x)\) doesnβt just give the value of the function for a specific
\(x\text{;}\) it also reminds us of the relationship between the input and output.
In the context of differential equations, function notation often appears alongside derivatives, such as
\(f'(x)\) or
\(\frac{dy}{dx}\text{.}\) These expressions indicate the rate of change of a function and are critical for describing real-world phenomena like motion, population growth, or heat transfer.
π Example 310. Simple Function Example.
Suppose
\(f(x) = 2x + 3\text{.}\) What is
\(f(4)\text{?}\)
To evaluate, substitute \(x = 4\) into the function:
\begin{equation*}
f(4) = 2(4) + 3 = 11.
\end{equation*}
This tells us that when the input is 4, the output of the function is 11.
Function notation also allows us to describe more complex relationships. For instance, in the differential equation
\(\frac{dy}{dx} = f(x) \cdot g(y)\text{,}\) we use functions
\(f(x)\) and
\(g(y)\) to represent how the rate of change of
\(y\) depends separately on
\(x\) and
\(y\text{.}\) Each function is flexible enough to represent anything from a simple polynomial to a more intricate expression involving trigonometric or exponential terms.
π Example 311. Function Composition and Relationships.
Consider two functions: \(f(x) = \sin(x)\) and \(g(y) = e^y\text{.}\) The composition of these functions, written as \(g(f(x))\text{,}\) represents the output of \(g\) when \(f(x)\) is used as its input. That is:
\begin{equation*}
g(f(x)) = e^{\sin(x)}.
\end{equation*}
This compact notation conveys a lot of information and is frequently used in modeling.
Function notation is especially powerful when combined with initial conditions in differential equations. For example, if we know that
\(\frac{dy}{dx} = 3x\) and
\(y(0) = 5\text{,}\) we can interpret
\(y(x)\) as a function that satisfies both the differential equation and the initial condition. Here,
\(y(0) = 5\) tells us that when
\(x = 0\text{,}\) the output of the function
\(y\) is 5.
π Example 312. Function Notation in Initial Value Problems.
Solve the initial value problem:
\begin{equation*}
\frac{dy}{dx} = 2x, \quad y(1) = 4.
\end{equation*}
Start by finding the general solution:
\begin{equation*}
y(x) = x^2 + C.
\end{equation*}
Then, use the initial condition \(y(1) = 4\) to determine \(C\text{:}\)
\begin{equation*}
4 = 1^2 + C \quad \Rightarrow \quad C = 3.
\end{equation*}
Thus, the particular solution is:
\begin{equation*}
y(x) = x^2 + 3.
\end{equation*}
To summarize, function notation is not just a way to write formulas, it is a powerful language for describing relationships, modeling systems, and solving problems. Mastering function notation ensures a smoother understanding of differential equations and opens the door to a deeper appreciation of how mathematics connects to the real world.
Function notation is a critical tool in mathematics, especially when working with differential equations. It allows us to concisely express relationships between variables, their rates of change, and how these relationships evolve over time or in different contexts. However, interpreting function notation requires careful attention to what each part of the equation represents.
Consider the following equation:
\begin{equation*}
\frac{d^n y}{dt^n} + a(t)y = k(t).
\end{equation*}
At first glance, this equation may seem intimidating, but it is simply a mathematical way to describe how a function \(y(t)\) behaves. Letβs break it down piece by piece.
Key Components of Function Notation.
1.
The Function: In this equation,
\(y = y(t)\) is a function of
\(t\text{,}\) meaning its value depends on the variable
\(t\text{.}\) For instance,
\(y(t)\) might represent a population at time
\(t\text{,}\) the temperature at a certain moment, or the position of an object.
2. Derivatives: The term \(\frac{d^n y}{dt^n}\) represents the \(n\)th derivative of \(y\) with respect to \(t\text{.}\) Derivatives measure how \(y(t)\) changes as \(t\) changes. For example:
3. Coefficients and Functions: The terms \(a(t)\) and \(k(t)\) are functions of \(t\text{.}\) They act as "modifiers" that influence the behavior of \(y(t)\text{.}\) For example:
These terms make the equation more flexible and realistic for modeling real-world systems.
Interpreting an Equation.
To better understand function notation, letβs examine an example:
\begin{equation*}
\frac{d^2y}{dt^2} + 5\frac{dy}{dt} + 6y = 10e^{-t}.
\end{equation*}
This is a second-order differential equation where:
-
\(\frac{d^2y}{dt^2}\) is the second derivative of \(y\text{,}\) representing acceleration or curvature in the system.
-
\(5\frac{dy}{dt}\) adds a damping term proportional to the first derivative (rate of change).
-
\(6y\) scales the function \(y(t)\) by a constant factor.
-
\(10e^{-t}\) is an external input that decays over time, influencing \(y(t)\text{.}\)
Together, these terms describe a system whose behavior depends on how \(y\) and its derivatives interact with both internal factors (like \(5\frac{dy}{dt}\)) and external forces (\(10e^{-t}\)).
When you encounter equations like this, focus on identifying the structure:
-
What is the highest-order derivative? This tells you the order of the equation.
-
Are there any coefficients (like \(5\) or \(6\)) or functions (like \(e^{-t}\)) that modify the behavior of \(y(t)\text{?}\)
-
Is there an external forcing term (e.g., \(k(t)\)) driving the system?
By answering these questions, you can build an intuition for how the equation models a real-world phenomenon.
π Example 313. Breaking Down an Equation.
Letβs analyze:
\begin{equation*}
\frac{dy}{dx} = 3x^2y.
\end{equation*}
This is a first-order differential equation where:
-
\(\frac{dy}{dx}\) represents how \(y\) changes with respect to \(x\text{.}\)
-
\(3x^2\) is a coefficient that depends on \(x\text{,}\) scaling \(y\text{.}\)
-
The product \(3x^2y\) shows that the rate of change of \(y\) depends on both \(x\) and the current value of \(y\text{.}\)
This form is separable, as we can rewrite it as:
\begin{equation*}
\frac{1}{y} \, dy = 3x^2 \, dx.
\end{equation*}
Recognizing this structure helps us know how to approach solving the equation.
Understanding function notation is key to reading, interpreting, and solving differential equations. By breaking down each term and identifying its role, you can gain a clear picture of how an equation models a systemβs behavior. Whether youβre dealing with a simple equation like
\(\frac{dy}{dx} = x\) or a complex one like
\(\frac{d^n y}{dt^n} + a(t)y = k(t)\text{,}\) function notation provides a powerful framework for understanding and solving mathematical problems.
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Section Solving Polynomial Equations
Higher-degree polynomial equations have the form
\begin{align*}
\text{degree 3:} \quad \amp a\ r^3 + b\ r + c\ r + d = 0, \\
\text{degree 4:} \quad \amp a\ r^4 + b\ r^3 + c\ r^2 + d\ r + e = 0, \\
\text{degree 5:} \quad \amp a\ r^5 + b\ r^4 + c\ r^3 + d\ r^2 + e\ r + f = 0, \\
\amp \vdots
\end{align*}
and it turns out that these equations can always be factored into simpler polynomials. In particular, a polynomial of degree \(n\) can always be factored into \(n\) linear factors:
\begin{align*}
\text{degree 3:} \quad \amp (r - r_1)(r - r_2)(r - r_3) = 0, \\
\text{degree 4:} \quad \amp (r - r_1)(r - r_2)(r - r_3)(r - r_4) = 0, \\
\text{degree 5:} \quad \amp (r - r_1)(r - r_2)(r - r_3)(r - r_4)(r - r_5) = 0, \\
\amp \vdots
\end{align*}
where \(r_1, r_2, r_3, ...\) are the solutions and can be real or complex. While knowing this is powerful, the process of factoring them can be quite challenging. However, there are some special forms and strategies that can help. A few are summarized below.
β³οΈ Techniques for Solving Higher-Degree Polynomials.
- Recognizing Special Forms
Some polynomials can be factored using special patterns. Common forms include:
-
Common Factoring: \(ax^n + bx^{n-1} = x^{n-1}(ax + b)\)
-
Difference of Squares: \(a^2 - b^2 = (a - b)(a + b)\)
-
Sum/Difference of Cubes: \(a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)\)
Recognizing these forms can help simplify the factorization process.
- Known Factors
If you know one factor, then you can βdivide-outβ the polynomial by this factor. For example, suppose we know that \(x = 1\) is a root of the polynomial
\begin{equation*}
x^3 - 6x^2 + 11x - 6 = 0\text{.}
\end{equation*}
Then, we know that \((x - 1)\) is a factor, so
\begin{equation*}
x^3 - 6x^2 + 11x - 6 = (x - 1)p(x)
\end{equation*}
where \(p(x)\) is some second-degree polynomial you multiply by \((x - 1)\) to get our original polynomial. We can find \(p(x)\) by dividing both sides by \(x - 1\text{,}\) like so
\begin{equation*}
p(x) = \frac{x^3 - 6x^2 + 11x - 6}{x - 1} \os{\knowl{./knowl/xref/poly-div.html}{how?}}{=} x^2 - 5x + 6
\end{equation*}
Therefore,
\begin{align*}
x^3 - 6x^2 + 11x - 6 \amp = (x - 1)(x^2 - 5x + 6)\\
\amp = (x - 1)(x - 2)(x - 3)
\end{align*}
- Possible Rational Roots
There is a theorem that tells us potential fractional roots of the polynomial. If \(r = \frac{p}{q}\) is a solution, then \(r\) must be of the form
\begin{equation*}
r = \pm \frac{\text{factor of } a_0}{\text{factor of } a_n}\text{.}
\end{equation*}
This gives us a list of possible solutions to test. For example, if we have the polynomial
\begin{equation*}
2x^3 - 3x^2 - 11x + 6 = 0
\end{equation*}
then the possible factors of \(a_0 = 6\) are \(1, 2, 3, 6\) and the possible factors of \(a_n = 2\) are \(1, 2\text{.}\) Therefore, the possible fractional solutions are
\begin{equation*}
\pm \frac{\text{factor of 6}}{\text{factor of 2}} = \pm \frac11, \pm \frac12, \pm \frac21, \pm \frac22, \pm \frac31, \pm \frac32, \pm \frac61, \pm \frac62
\end{equation*}
We can test each of these values to find up to 3 solutions. Once we find one, we can use the previous technique to help find more.
- Use Technology
Factoring higher-order polynomials can be very difficult to do by and this is one skill that may be better suited for a computer. There are many software packages that can factor polynomials for you. For example, the Wolfram Alpha website will do it with ease.
π Example 314.
Completely factor and solve the following characteristic equations
-
\(\displaystyle 4\ r^2 - 9 = 0 \)
-
\(\displaystyle r^2 - 3 = 0 \)
-
\(\displaystyle r^3 + 3\ r^2 - 4\ r = 0 \)
-
\(\displaystyle r^5 + 10\ r^4 = 0 \)
-
\(\displaystyle r^4 - 100\ r^2 = 0 \)
-
\(\displaystyle r^5 - 4\ r^3 = 0 \)
-
\(\displaystyle r^4 - 32 = 0 \)
Solution.
The degree of each equation tells you how many factors you should have.
-
| \(4\ r^2 - 9 = 0 \) |
|
| \((2\ r + 3)(2\ r - 3) = 0 \) |
\(\leftarrow\) difference of squares |
| \(2\ r + 3 = 0, \quad 2\ r - 3 = 0 \) |
\(\leftarrow\) set each factor to 0 |
| \(\ds r_1 = -\frac32, \quad r_2 = \frac32 \) |
\(\leftarrow\) solutions |
-
| \(r^2 - 3 = 0 \) |
|
| \((r + \sqrt{3})(r - \sqrt{3}) = 0 \) |
\(\leftarrow\) difference of squares |
| \(\ds r_1 = -\sqrt{3}, \quad r_2 = \sqrt{3} \) |
\(\leftarrow\) solutions |
-
| \(r^3 + 3\ r^2 - 4\ r = 0 \) |
|
| \(r\ (r^2 + 3\ r - 4) = 0 \) |
\(\leftarrow\) common factor |
| \(r\ (r + 4)(r - 1) = 0 \) |
\(\leftarrow\) standard quadratic factoring |
| \(\ds r_1 = 0, \quad r_2 = -4, \quad r_3 = 1 \) |
\(\leftarrow\) solutions |
-
| \(r^5 + 10\ r^4 = 0 \) |
|
| \(r^4\ (r + 10) = 0 \) |
\(\leftarrow\) common factor |
| \(\ds r_1 = 0\ (\text{4 repeats}), \quad r_2 = -10 \) |
\(\leftarrow\) solutions |
Technically, \(r^4 = (r-0)^4\) and represents 4 repeated factors.
-
| \(r^4 - 100\ r^2 = 0 \) |
|
| \(r^2\ (r^2 - 100) = 0 \) |
\(\leftarrow\) common factor |
| \(r^2\ (r + 10)(r - 10) = 0 \) |
\(\leftarrow\) difference of squares |
| \(\ds r_1 = 0\ (\text{2 repeats}), \quad r_2 = -10, \quad r_3 = 10 \) |
\(\leftarrow\) solutions |
-
| \(r^5 - 4\ r^3 = 0 \) |
|
| \(r^3\ (r^2 - 4) = 0 \) |
\(\leftarrow\) common factor |
| \(r^3\ (r + 2)(r - 2) = 0 \) |
\(\leftarrow\) difference of squares |
| \(\ds r_1 = 0\ (\text{3 repeats}), \quad r_2 = -2, \quad r_3 = 2 \) |
\(\leftarrow\) solutions |
-
| \(r^4 - 25 = 0 \) |
|
| \((r^2 + 5)(r^2 - 5) = 0 \) |
\(\leftarrow\) difference of squares |
| \((r^2 + 5)(r - \sqrt{5})(r + \sqrt{5}) = 0 \) |
\(\leftarrow\) difference of squares |
| \(r^2 + 5 = 0, \quad r - \sqrt{5} = 0, \quad r + \sqrt{5} = 0 \) |
\(\leftarrow\) set each factor to 0 |
| \(\ds r_1 = -i\sqrt{5}, \quad r_2 = i\sqrt{5}, \quad r_3 = -\sqrt{5}, \quad r_4 = \sqrt{5} \) |
\(\leftarrow\) solutions |
An important concept to remember is that any polynomial can be factored into the product of linear factors, allowing for complex solutions. This is known as the Fundamental Theorem of Algebra. However, factoring higher-degree polynomials can sometimes be challenging and may require the use of technology, such as computer algebra systems or graphing calculators, to find complex or irrational roots.
By combining these techniques, we can solve for the roots of any higher-degree polynomial. Once we have the roots, we can construct the general solution to the higher-order LHCC equation.
For a higher-order LHCC equation like:
\begin{equation}
a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y' + a_0 y = 0\text{,}\tag{49}
\end{equation}
the characteristic equation is the polynomial equation we just discussed. Finding the roots \(r_1, r_2, \ldots, r_n\) of this polynomial gives us the general solution:
\begin{equation*}
y = C_1 e^{r_1 x} + C_2 e^{r_2 x} + \cdots + C_n e^{r_n x}\text{,}
\end{equation*}
where \(C_1, C_2, \ldots, C_n\) are constants determined by initial conditions.
Letβs see an example to solidify these concepts.
π Example 315. Solving a Third-Degree Polynomial Equation.
Find the general solution to the third-order LHCC equation:
\(y''' - 6y'' + 11y' - 6y = 0 \)
Solution.
First, write down the characteristic equation:
\begin{gather*}
r^3 - 6r^2 + 11r - 6 = 0 \text{.}
\end{gather*}
Factoring the polynomial, we get:
\begin{equation*}
(r - 1)(r - 2)(r - 3) = 0\text{.}
\end{equation*}
Therefore, the roots are \(r = 1, 2, 3\text{.}\)
Since we have three distinct real roots, the general solution to the LHCC equation is:
\begin{equation*}
y = C_1 e^{x} + C_2 e^{2x} + C_3 e^{3x}\text{,}
\end{equation*}
where \(C_1, C_2, C_3\) are constants determined by initial conditions.
Polynomail Division.
Use polynomial divison to compute
\begin{equation*}
\frac{x^3 - 6x^2 + 11x - 6}{x-1} =\ ?
\end{equation*}
|
|
|
\(x^2\) |
\(-\) |
\(5x\) |
\(+\) |
\(6\) |
| \(x-1\) |
\(x^3\) |
\(-\) |
\(6x^2\) |
\(+\) |
\(11x\) |
\(-\) |
\(6\) |
|
\(x^3\) |
\(-\) |
\(x^2\) |
|
|
|
|
|
|
|
\(-5x^2\) |
\(+\) |
\(11x\) |
\(-\) |
\(6\) |
|
|
|
\(-5x^2\) |
\(+\) |
\(5x\) |
|
|
|
|
|
|
|
\(6x\) |
\(-\) |
\(6\) |
|
|
|
|
|
\(6x\) |
\(-\) |
\(6\) |
|
|
|
|
|
|
|
\(0\) |
Therefore,
\begin{equation*}
\frac{x^3 - 6x^2 + 11x - 6}{x-1} = x^2 - 5x + 6\text{.}
\end{equation*}
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Section Piecewise Defined Functions
We will encounter piecewise defined functions in differential equations when we think about some physical phenomenon. For example, we might consider the vibration of an airplane wing that is struck by some external object or a circuit that is initially open and then we close the circuit and the current immediately starts flowing. Both of these scenarios would require a piecewise defined function because there is a moment in time when something about the physical system changes.
As such, we need to remember how piecewise defined functions work. Consider the function
\begin{equation*}
g(t) = \left\{
\begin{array}{ll}
0, \amp t \lt 1 \\
t^2, \amp 1 \le t \lt 2 \\
-t+1, \amp 2 \le t \lt 6 \\
0, \amp t \ge 6. \\
\end{array} \right.
\end{equation*}
If we want to evaluate the function at a particular
\(t\)-value, we use the restrictions on the right to point us to which piece of the function definition we should use. For example, if we want to know
\(g(3)\text{,}\) then we look over at that right side and see that 3 falls into the interval
\(2 \le t \lt 6,\) so we use the corresponding function,
\(-t+1\text{.}\) Thus,
\(g(3) = -3 + 1 = -2.\)
If we want to graph, we also use those restrictions. When
\(t \lt 1\text{,}\) or equivalently when
\(t\) is in the interval
\((-\infty, 1),\) we know that the graph of
\(g\) will look like the horizontal line
\(y=0\text{;}\) on the interval
\([1, 2),\) the graph will look like the graph of the parabola
\(y = t^2\text{;}\) etc.
In the plots below you can see the entire graphs of the functions, dotted, with a solid segment that will be part of the piecewise function
\(g\text{.}\)
Now we are prepared to assemble the pieces and generate the graph of the piecewise-defined function
\begin{equation*}
g(t) = \left\{
\begin{array}{ll}
0, \amp t \lt 1 \\
t^2, \amp 1 \le t \lt 2 \\
-t+1, \amp 2 \le t \lt 6 \\
0, \amp t \ge 6 \\
\end{array} \right.\text{.}
\end{equation*}
If youβd like more of a review feel free to look at VMIβs precalculus text,
here
Sketch each of the following piecewise defined functions.
-
\(\displaystyle \ds f(t) = \left\{
\begin{array}{ll}
0, \amp t \lt 0 \\
5-t, \amp 0 \le t \lt 3 \\
2, \amp t \ge 3 \\
\end{array} \right.\)
Answer.
-
\(\displaystyle \ds h(t) = \left\{
\begin{array}{ll}
\sin t, \amp t \lt \pi \\
0, \amp t \ge \pi \\
\end{array} \right.\)
Answer.
-
\(\displaystyle \ds w(t) = \left\{
\begin{array}{ll}
-t, \amp t \lt 1 \\
t^2+1, \amp 1 \le t \lt 2 \\
e^{-t}, \amp t \ge 2 \\
\end{array} \right.\)
Answer.
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