DE-BOOK Quick References

Section Quick References

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Subsubsection Basics

Worksheet Key Terms & Concepts

✳️ Differential Equations and their Components.

Differential Equation (DE) 🔗: An equation that involves one or more derivatives of an unknown function.
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Variables 🔗: The dependent variable represents the unknown function that you are solving for and always has derivatives applied to it. The dependent variable is a function of the independent variable. In

\begin{gather*} 12y'' + (x+5)y' - \ln(x) y = 5 - \cos x\text{,} \end{gather*}

dependent \(\leftarrow y\quad\) & \(\quad\) independent \(\leftarrow x\text{.}\)

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Terms & Free Terms 🔗: Parts of an equation separated by \(+\text{,}\) \(-\text{,}\) or \(=\) and each containing a different form of the dependent variable. The collection of all terms without a dependent variable is referred to as a free term.
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\begin{gather*} \us{\large y'' \text{ term}}{\boxed{12y''}} + \us{\large y' \text{ term}}{\boxed{(x+5)y'}} - \us{\large y \text{ term}}{\boxed{\ln(x) y}} = \us{\large\text{free term}}{\boxed{5 - \cos x}} \end{gather*}
Coefficients 🔗: The part of a term multiplied by the dependent variable or its derivatives.
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\begin{gather*} \boxed{12}\us{\Large y'',\ y',\ y \text{ coefficients}}{\us{\nwarrow}{\ y}'' + \us{\uparrow}{\boxed{(x+5)}}\ y' \us{\nearrow}{-\ }} \boxed{\ln (x)}\ y = 5 - \cos x \end{gather*}

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✳️ Order & Linearity.

Order 🔗: The highest order derivative present in a DE.
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Linear Term 🔗: A term of the form:
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\begin{equation*} a(t)\ y,\ a(t)\ y',\ a(t)\ y'',\ a(t)\ y''',\ \ldots\text{,} \end{equation*}

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where \(y\) is the dependent variable, and \(a(t)\) is a coefficient that depends only on the independent variable \(t\text{.}\)
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Linear DE 🔗: A DE composed entirely of linear terms.
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Nonlinear DE 🔗: A DE that contains at least one nonlinear term.
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Subsubsection Solutions to Differential Equations

Worksheet Key Terms & Concepts

✳️ Solutions & Initial Conditions.

Satisfying a DE 🔗: A function satisfies a DE if substituting it into the dependent variable results in the equation simplifying to a true statement (e.g., \(0 = 0\)).
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🔗: A function that satisfies the DE.
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General Solution 🔗: The common form (template) of all the solutions in the family. It contains constants that can take any value.
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Particular Solution 🔗: A single solution obtained by assigning specific values to the constants in the general solution.
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Family of Solutions 🔗: The collection of all possible particular solutions.
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Initial Conditions 🔗: Known values of the solution or its derivatives at a specific point, used to determine a particular solution from the general solution.
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✳️ Direct Integration.

Direct Integration 🔗: A method to solve differential equations of the form:

\begin{equation*} \frac{d}{dx}\left[g(x,y)\right] = f(x), \end{equation*}

by integrating both sides with respect to the independent variable \(x\text{.}\)

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Subsubsection Separation of Variables

Worksheet Key Terms & Concepts

✳️ First-Order Differential Equations.

First-Order Differential Equation 🔗: Every first-order differential equation can be written in the form

\begin{equation*} f(x, y, y') = 0\text{.} \end{equation*}

This just means that all the terms in the equation have been moved to the left-hand side, which can only contain the independent variable, \(x\text{,}\) the dependent variable, \(y\text{,}\) and its derivative, \(y'\text{.}\)

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✳️ Separation of Variables.

First-Order Separable Differential Equation 🔗: These equations can be written as:

\begin{equation*} \frac{dy}{dx} = f(x) \cdot g(y). \end{equation*}

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Separation of Variables Method 🔗: A solution technique for separable equations. It involves isolating \(y\) and \(x\) terms on opposite sides, followed by integration.
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Subsubsection Integrating Factor

Worksheet Key Terms & Concepts

✳️ Integrating Factor.

First-Order Linear Differential Equation 🔗: These equations take the standard form:

\begin{equation*} y' + P(x)y = Q(x). \end{equation*}

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Integrating Factor 🔗: A function, \(\mu\text{,}\) multiplied onto the standard form, above, to reverse the product rule, leading to the equation, \(\mu' = P\mu\text{,}\) with the solution

\begin{equation*} \mu = e^{\int P(x) dx}. \end{equation*}

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Integrating Factor Method 🔗: A solution method for first-order linear equations that uses an integrating factor to convert the equation into a form solvable by direct integration.
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Subsubsection Qualitative Methods

Worksheet Key Terms & Concepts

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Subsubsection Numerical Methods

Worksheet Key Terms & Concepts

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Subsubsection Linear Homogeneous Differential Equations with Constant Coefficients

The LHCC chapter (Linear Homogeneous Differential Equations with Constant Coefficients) in "Linear Constant Coefficient Methods" introduces a systematic method to solve higher-order linear differential equations. Here’s a summary based on the content:

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Worksheet Key Terms & Concepts

✳️ Summary of the Key Ideas.

Linear Homogeneous Differential Equations with Constant Coefficients (LHCC)
- These are differential equations where each term consists of a derivative of the unknown function multiplied by a constant.
  
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- The general form of an LHCC equation is:
  \begin{equation*} a_n\ y^{(n)} + a_{n-1}\ y^{(n-1)} + \dots + a_1\ y' + a_0\ y = 0\text{.} \end{equation*}
  
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The Characteristic Equation
- By assuming a solution of the form \(y = e^{rx}\text{,}\) an LHCC can be reduced to a characteristic polynomial in \(r\text{.}\)
  
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- The solutions to the characteristic equation determine the form of the general solution.
  
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Solution Types
- Let \(r\) be a solution to the characteristic equation (CE).
  
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- If \(r\) is different from all other solutions of the CE, then
  \begin{equation*} c e^{r x} \end{equation*}
  is a term of the general solution.
  
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- If \(r\) is equal to, say, three other solutions of the CE, then
  \begin{equation*} c_1 e^{r x} + c_2 x e^{r x} + c_3 x^2 e^{r x} \end{equation*}
  are terms of the general solution.
  
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- If \(r = \alpha + i\beta\) or \(r = \alpha - i\beta\text{,}\) then the general solution contains
  \begin{equation*} e^{\alpha x}(c_1\sin(\beta x)+c_2\cos(\beta x))\text{.} \end{equation*}
  
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✍🏻 Method 7. LHCC Method.

The general solution to a linear homogeneous differential equation with constant coefficients (LHCC) of the form

\begin{equation} a_n\ y^{(n)} + a_{n-1}\ y^{(n-1)} + \cdots + a_2\ y'' + a_1\ y' + a_0\ y = 0,\tag{51} \end{equation}

can be found through the following steps...

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Step 1: Solve the Characteristic Equation 🔗

Solve the characteristic equation (CE)

\begin{equation*} a_n\ r^{n} + a_{n-1}\ r^{n-1} + \cdots + a_2\ r^2 + a_1\ r + a_0 = 0, \end{equation*}

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Step 2: Write Down the General Solution 🔗

Real & Different: \(r_1, r_2, \dots, r_n \)
\begin{equation*} y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x} + \dots + c_n e^{r_n x}\text{.} \end{equation*}

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Real & Repeated: \(r_1 \) (multiplicity \(m \))
\begin{equation*} y(x) = (c_1 + c_2 x + \dots + c_m x^{m-1}) e^{r_1 x}\text{.} \end{equation*}

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Complex: \(\alpha \pm i\beta \)
\begin{equation*} y(x) = e^{\alpha x} \left(c_1 \cos(\beta x) + c_2 \sin(\beta x)\right)\text{.} \end{equation*}

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For mixed root types, combine the corresponding terms to form the complete general solution.

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Substituting \(y = e^{rx}\) transforms the differential equation into an algebraic equation for \(r\text{.}\)

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The characteristic equation is a polynomial equation whose roots determine the fundamental solutions.

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An n-th order LHCC equation always has n solutions, corresponding to the n roots of its characteristic equation.

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Each real root \(r\) leads to a solution of the form \(e^{rx}\text{.}\)

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In the next subsection, we will explore how to construct the general solution by combining these exponential solutions appropriately.

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The key ideas are:

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The superposition principle allows us to sum independent solutions to form the general solution.

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The number of fundamental solutions matches the order of the differential equation.

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The general solution consists of all fundamental solutions, each multiplied by an arbitrary constant.

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✳️ Properties of LHCC Solutions.

The solutions to any \(n\)-th order linear homogeneous differential equation,

\begin{equation} a_n\ y^{(n)} + \cdots + a_2\ y'' + a_1\ y' + a_0\ y = 0\tag{52} \end{equation}

have the following properties:

Exponential Solutions 🔗: The order of equation (52) tells us that it has exactly \(n\) solutions, each assuming the form,

\begin{equation} e^{rx} \quad \text{or} \quad x^k e^{rx}\text{,}\tag{53} \end{equation}

where \(r\) and \(k\) are numbers specific to this equation.

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Linearly Independent 🔗: The \(n\) solutions to an \(n\)-th order LHCC equation are linearly independent, meaning no two solutions can be combined into a single term.
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General Solution 🔗: Let \(y_1, y_2, \dots, y_n\) be the \(n\) linearly independent exponential solutions, then the general solution is formed by

\begin{equation*} y = c_1 y_1 + c_2 y_2 + \cdots + c_n y_n \end{equation*}

for any constants \(c_1, c_2, \dots, c_n\text{.}\)

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✳️ Key Takeaways.

If \(y_1\) and \(y_2\) are solutions to an LHCC equation, then \(c_1 y_1 + c_2 y_2\) is also a solution for any constants \(c_1\) and \(c_2\text{.}\)

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The general solution to an LHCC equation consists of a linear combination of all its independent solutions.

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The number of independent solutions corresponds to the order of the equation.

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An LHCC equation often has multiple exponential solutions, determined by solving the characteristic equation.

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The principle of superposition states that any linear combination of these solutions is also a solution.

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The general solution is formed by summing all independent exponential solutions with arbitrary coefficients.

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The solutions to LHCC equations are exponential functions because their derivatives preserve the same functional form.

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The concept of “like-terms” helps explain why other function types (polynomials, trigonometric functions) do not work.

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By substituting an exponential \(y = e^{rx}\) into the equation, we obtain a polynomial equation for \(r\text{,}\) which determines the general solution.

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In this subsection, we developed a framework for solving Linear Homogeneous Constant Coefficient (LHCC) equations. We began by exploring why exponential functions serve as the fundamental solutions to these equations, using the concept of like-terms to justify their role.

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We then introduced the characteristic equation, which transforms the problem of solving a differential equation into solving a polynomial equation. This method allows us to systematically determine the correct set of fundamental solutions for any LHCC equation.

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With this foundation in place, we are now ready to introduce a powerful tool for systematically finding these solutions: the characteristic equation. This algebraic approach allows us to determine the exponential solutions efficiently and will serve as the backbone of our solution methods moving forward.

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To build intuition, we will carefully examine the three defining properties of LHCC equations: linearity, homogeneity, and constant coefficients. Each of these characteristics influences the types of solutions that emerge and dictates the methods we use to solve them. By clearly defining and identifying LHCC equations, we lay the groundwork for developing powerful solution techniques in the sections that follow.

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Unlike many other types of differential equations, LHCC equations have a highly structured solution method. Their solutions take on a predictable form, allowing us to develop a systematic approach for solving them. Through examples and analysis, we will discover that exponential functions play a central role in solving LHCC equations. By assuming solutions of the form \(y = e^{rx}\) and substituting into the equation, we obtain an algebraic equation for \(r\text{,}\) the characteristic equation. The solutions to this characteristic equation directly determine the general solution to the differential equation.

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Table 327. Examples of LHCC General Solutions

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Characteristic Equation Solutions	General Solution
\(r = 3, -3, 5.3\) (3rd order)	\begin{equation} c_1e^{3t} + c_2e^{-3t} + c_3e^{5.3t} \end{equation}
\(r = 6 \pm i\sqrt{7.7}, 0\) (3rd order)	\begin{equation} e^{6t}\left(c_1\sin(\sqrt{7.7}t) + c_2\cos(\sqrt{7.7}t)\right) + c_3 \end{equation}
\(r = -4 \text{ (triple)}, 5.3\) (4th order)	\begin{equation} (c_1t^2 + c_2t + c_3)e^{-4t} + c_4e^{5.3t} \end{equation}
\(r = \pm \frac{i}{2}, 2 \pm i\) (4th order)	\begin{equation} c_1\sin\left(\frac{t}{2}\right) + c_2\cos\left(\frac{t}{2}\right) + e^{2t}(c_3\sin t + c_4\cos t) \end{equation}
\(r = 0 \text{ (double)}, 3 \text{ (5-repeats)}\) (7th order)	\begin{equation} c_1t + c_2 + (c_3t^4 + c_4t^3 + c_5t^2 + c_6t + c_7)e^{3t} \end{equation}
\(r = \pm i, \pi \text{ (double)}, 5\) (5th order)	\begin{equation} c_1\sin t + c_2\cos t + (c_3t + c_4)e^{\pi t} + c_5e^{5t} \end{equation}

The key takeaways from this subsection are:

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Exponential functions naturally emerge as solutions to LHCC equations due to their unique differentiation properties.

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The characteristic equation provides a systematic way to determine the fundamental solutions.

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The general solution consists of all independent exponential solutions, each multiplied by an arbitrary constant.

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The number of solutions matches the order of the differential equation.

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In summary, a differential equation is classified as an LHCC equation if it satisfies three criteria:

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It is linear: each term involves \(y\) or its derivatives to the first power only.

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It is homogeneous: the right-hand side is zero.

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It has constant coefficients: the coefficients of \(y\) and its derivatives do not depend on \(x\text{.}\)

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Subsubsection Undermined Coefficients

Worksheet Key Terms & Concepts

✳️ Summary of the Key Ideas.

The method of undetermined coefficients is used to solve non-homogeneous linear differential equations.

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The general solution to a non-homogeneous equation is the sum of the general solution to the corresponding homogeneous equation and a particular solution.

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The method involves guessing the form of the particular solution based on the form of the non-homogeneous term and solving for the coefficients.

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The method is applicable when the non-homogeneous term can be expressed as a linear combination of known functions.

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LHCC Equation 🔗: An LHCC equation is a Linear Homogeneous Differential Equations with Constant Coefficients and have the form

\begin{equation*} a_n y^{(n)} + \cdots + a_1 y' + a_0 y = 0\text{.} \end{equation*}

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Characteristic Equation 🔗: 🔗
LHCC General Solutions 🔗: 🔗

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LNCC Equation 🔗: An LNCC equation is a Linear Nonhomogeneous Differential Equations with Constant Coefficients and have the form

\begin{equation} a_n y^{(n)} + \cdots + a_1 y' + a_0 y = f(x)\text{.}\tag{54} \end{equation}

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Forcing Function 🔗: A forcing function is the free term in (54), denoted by \(f(x)\text{.}\)
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Particular Solution 🔗: A particular solution, \(y_h\text{,}\) is a function that can be plugged into (54) and all the terms combine to give the forcing function, \(f(x)\text{.}\)
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Homogeneous Solution 🔗: A homogeneous solution, \(y_h\text{,}\) is a function such that, when it is plugged into (54) all the terms cancel to zero. That is, \(y_h\) is the solution to the homogeneous version of (54) (\(f(x) = 0 \)).
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General Solutions 🔗: The general solution to (54) is the sum of the homogeneous and particular solutions, that is

\begin{equation*} y = y_h + y_p\text{.} \end{equation*}

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	\(y\)	\(y'' - 4y' + 3{\color{blue} y} =\!\!\!\)	LHS	LHS \(\os{?}{=} 9x\)
1	\(3\)	\((3)'' - 4(3)' + 3({\color{blue} 3}) =\!\!\!\)	\({\color{blue} 9}\)	No
2	\(3x\)	\((3x)'' - 4(3x)' + 3({\color{blue} 3x}) =\!\!\!\)	\(-12 + {\color{blue} 9x}\)	No
3	\(x^4\)	\((x^4)'' - 4(x^4)' + 3({\color{blue} x^4}) =\!\!\!\)	\(12x^2 - 16x^3 + {\color{blue} 3x^4}\)	No
4	\(x^2+3x\)	\((x^2+3x)'' - 4(x^2+3x)' + 3({\color{blue} x^2}+3x) =\!\!\!\)	\(-10 + x + {\color{blue} 3x^2}\)	No
5	\(3x-6\)	\((3x-6)'' - 4(3x-6)' + 3({\color{blue} 3x}-6) =\!\!\!\)	\(-30+{\color{blue} 9x}\)	No
6	\(3x+4\)	\((3x+4)'' - 4(3x+4)' + 3({\color{blue} 3x}+4) =\!\!\!\)	\({\color{blue} 9x}\)	Yes

Based on this table, we note that the correct solution is \(y = 3x + 4\) and

Row 1 shows that a free term alone could never produce an \(x\) term.
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Derivatives reduce the power of a polynomial, so the highest power term (highlighted in blue) comes from the \(y\) term.
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\(y'' - 4y' + \os{\large y\text{ term}}{\boxed{3y}}\)
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Rows 3 & 4 illustrate the solution can’t have a \(x^2\) or higher-degree term.
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Row 6 shows the solution (\(3x+4\)) needed an \(x\) term and free term even though the right-hand side, \(9x\text{,}\) has only an \(x\) term.
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Subsection LNCC Equations

Figure 328. Comparison of a homogeneous equation (*left*) and a nonhomogeneous equation (*right*). In both, the solutions must simplify in a specific way when substituted into the equation.
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Table 329. row 1

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	\(f(x)\ \) type	\(y_p\) Form
1	\(a\) (constant)	\(A\)

Table 330. row 2

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	\(f(x)\ \) type	\(y_p\) Form
2	\(ax + b\)	\(Ax + B\)

Table 331. row 3

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	\(f(x)\ \) type	\(y_p\) Form
3	\(ax^2 + bx + c\)	\(Ax^2 + Bx + C\)

Table 332. row 4

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	\(f(x)\ \) type	\(y_p\) Form
4	\(ax^3 + bx^2 + cx + d\)	\(Ax^3 + Bx^2 + Cx + D\)

Table 333. row 5

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	\(f(x)\ \) type	\(y_p\) Form
5	\(a e^{\ds\alpha x}\)	\(A e^{\ds\alpha x}\)

Table 334. row 6

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	\(f(x)\ \) type	\(y_p\) Form
6	\(a \sin(\beta x) + b \cos(\beta x)\)	\(A \sin(\beta x) + B \cos(\beta x)\)

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Subsubsection Laplace Transforms

Worksheet Key Terms & Concepts

In this section, we introduced the concept of the forward Laplace transform and derived some common Laplace transforms that we will use throughtout this chapter. The following points summarize the essential concepts from the forward Laplace transform section:

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✳️ Summary of the Key Ideas.

Differential \(\to\) Algebraic Equations. The Laplace transform converts a differential equation into an algebraic equation, simplifying the solution process by eliminating derivatives.

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Laplace Transform Concept. Applying the Laplace transform to a differential equation involves transforming each term by multiplying by \(e^{-st}\) and integrating with respect to \(t\) from \(0\) to \(\infty\text{,}\) but is often simplified by directly applying the Laplace operator, \(\laplacesym\text{.}\)

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Linearity Property. The Laplace transform is linear, meaning it distributes across addition and subtraction, and allows for constants to be factored out. This property is essential for transforming complex equations.

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Transforming Initial Conditions. Initial conditions are incorporated directly into the Laplace-transformed equation, modifying the transformed terms to include initial values, making it easier to solve the resulting algebraic equation.

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Common Function Transforms. The Laplace transforms of common functions, such as exponentials, sines, cosines, and polynomials, are essential tools in transforming differential equations and are summarized in the provided table.

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Transforming Derivatives. The Laplace transform of a derivative, \(y'(t)\) or higher, transfers the derivative onto the Laplace variable \(s\text{,}\) reducing the order of the equation while introducing initial condition terms.

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Multiplication by \(e^{at}\) and \(t^n\). When multiplying a function by an exponential \(e^{at}\text{,}\) the Laplace transform shifts by \(a\) in the \(s\)-domain, and multiplying by \(t^n\) corresponds to differentiating the transform \(n\) times with respect to \(s\text{,}\) introducing a sign change.

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Transforming the Entire Equation. The process of applying the Laplace transform to an entire differential equation with initial conditions involves systematically transforming each term and leads to a simplified algebraic equation in the \(s\)-domain, ready for solving.

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Assume \(F(s) = \lap{f(t)}\text{,}\) \(a\text{,}\) \(b\) are constant, and \(n=0,1,2,3,\ldots\)

L\(_3\)\(\ds\quad\lap{ t^n } = \frac{n!}{s^{n+1}}, \quad s >0, \quad n = 1, 2, 3, \ldots \).

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L\(_4\)\(\ds\quad\lap{ \sin(bt) } = \frac{b}{s^2+b^2}, \quad s >0\).

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L\(_5\)\(\ds\quad\lap{ \cos(bt) } = \frac{s}{s^2+b^2}, \quad s >0\).

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L\(_6\)\(\ds\quad\lap{ e^{at}\ t^n } = \frac{n!}{(s-a)^{n+1}}, \quad s >a, \quad n = 0, 1, 2, 3, \ldots \).

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L\(_7\)\(\ds\quad\lap{ e^{at}\sin(bt) } = \frac{b}{(s-a)^2+b^2}, \quad s >a\).

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L\(_8\)\(\ds\quad\lap{ e^{at}\cos(bt) } = \frac{s-a}{(s-a)^2+b^2}, \quad s >a\).

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Laplace Transform of Derivatives.

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L\(_9\)\(\ds\quad\lap{ f'(t) } = sF(s) - f(0), \quad s > 0\).

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L\(_{10}\)\(\ds\quad\lap{ f''(t) } = s^2F(s) - sf(0) - f'(0), \quad s > 0\).

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L\(_{11}\)\(\ds\quad\lap{ f'''(t) } = s^3F(s) - s^2f(0) - sf'(0) - f''(0), \quad s > 0\).

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L\(_{12}\)\(\ds\quad\lap{ t^n f(t) } = (-1)^n \frac{d^n}{ds^n}F(s), \quad s > 0, \quad n = 0, 1, 2, 3, \ldots \).

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Table 335. Common Laplace Transforms. \(a, b\) are constant, \(n = 1, 2, \ldots\)

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	\(t\)-functions	\(s\)-functions
	\(\downarrow\)	\(\downarrow\)
	\(f(t)\)	\(\lap{ f(t) }\)
\(L_1\)	\(1\)	\(\ds \frac{1}{s}\)	\(s \gt 0\)
\(L_2\)	\(e^{at}\)	\(\ds \frac{1}{s-a}\)	\(s \gt a\)
\(L_3\)	\(t^n\)	\(\ds \frac{n!}{s^{n+1}}\)	\(s \gt 0\)
\(L_4\)	\(\sin (bt)\)	\(\ds \frac{b}{s^2 + b^2}\)	\(s \gt 0\)
\(L_5\)	\(\cos(bt)\)	\(\ds \frac{s}{s^2 + b^2}\)	\(s \gt 0\)
\(L_6\)	\(t^n e^{at}\)	\(\ds \frac{n!}{(s-a)^{n+1}}\)	\(s \gt a\)
\(L_7\)	\(e^{at} \sin(bt)\)	\(\ds \frac{b}{(s-a)^2 + b^2}\)	\(s \gt a\)
\(L_8\)	\(e^{at} \cos(bt)\)	\(\ds \frac{s-a}{(s-a)^2 + b^2}\)	\(s \gt a\)

Table 336. Laplace Transforms Properties, \({\small \lap{ y(t) } = Y(t), \ \lap{ z(t) } = Z(t)}\)

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	Properties
\(P_1\)	\(\ds \lap{ a y(t) \pm b z(t) } = a Y(s) \pm b Z(s) \vphantom{\frac{d}{d}}\)
\(P_2\)	\(\ds \lap{ e^{at} y(t) } = Y(s-a) \vphantom{\frac{d}{d}}\)
\(P_3\)	\(\ds \lap{ y'(t) } = s\,Y(s) - y(0) \vphantom{\frac{d}{d}}\)
\(P_4\)	\(\ds \lap{ y''(t) } = s^2\,Y(s) - s\,y(0) - y'(0) \vphantom{\frac{d}{d}}\)
\(P_5\)	\(\ds \lap{ y'''(t) } = s^3\,Y(s) - s^2\,y(0) - s\,y'(0) - y''(0) \vphantom{\frac{d}{d}}\)
\(P_6\)	\(\ds \lap{ t^n y(t) } = (-1)^n \frac{d^{(n)}}{ds^{(n)}}Y(s) \vphantom{\frac{d}{d}}\)

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Table 337. Table of Common Laplace Transforms

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	Function (\(t\)-Domain) \begin{equation} f(t) \end{equation}	Laplace Transform (\(s\)-Domain) \begin{equation} \lap{f(t)} = F(s) \end{equation}	Existence Condition
L\(_1\)	\begin{equation} 1 \end{equation}	\begin{equation} \frac{1}{s} \end{equation}	\begin{equation} s > 0 \end{equation}
L\(_2\)	\begin{equation} e^{at} \end{equation}	\begin{equation} \frac{1}{s - a} \end{equation}	\begin{equation} s > a \end{equation}
L\(_3\)	\begin{equation} t^n \end{equation}	\begin{equation} \frac{n!}{s^{n+1}} \end{equation}	\begin{equation} s > 0 \end{equation}
L\(_4\)	\begin{equation} \sin(bt) \end{equation}	\begin{equation} \frac{b}{s^2+b^2} \end{equation}	\begin{equation} s > 0 \end{equation}
L\(_5\)	\begin{equation} \cos(bt) \end{equation}	\begin{equation} \frac{s}{s^2+b^2} \end{equation}	\begin{equation} s > 0 \end{equation}
L\(_6\)	\begin{equation} e^{at}\ t^n \end{equation}	\begin{equation} \frac{n!}{(s-a)^{n+1}} \end{equation}	\begin{equation} s > a \end{equation}
L\(_7\)	\begin{equation} e^{at}\sin(bt) \end{equation}	\begin{equation} \frac{b}{(s-a)^2+b^2} \end{equation}	\begin{equation} s > a \end{equation}
L\(_8\)	\begin{equation} e^{at}\cos(bt) \end{equation}	\begin{equation} \frac{s-a}{(s-a)^2+b^2} \end{equation}	\begin{equation} s > a \end{equation}
R\(_{1}\)	\begin{equation} f'(t) \end{equation}	\begin{equation} sF(s) - f(0) \end{equation}	\begin{equation} s > 0 \end{equation}
R\(_{2}\)	\begin{equation} f''(t) \end{equation}	\begin{equation} s^2F(s) - sf(0) - f'(0) \end{equation}	\begin{equation} s > 0 \end{equation}
R\(_{3}\)	\begin{equation} f'''(t) \end{equation}	\begin{equation} s^3F(s) - s^2f(0) - sf'(0) - f''(0) \end{equation}	\begin{equation} s > 0 \end{equation}
R\(_{4}\)	\begin{equation} e^{at} f(t) \end{equation}	\begin{equation} F(s-a) \end{equation}	\begin{equation} s > 0 \end{equation}
R\(_{5}\)	\begin{equation} t^n f(t) \end{equation}	\begin{equation} (-1)^n \frac{d^n}{ds^n}\Big[F(s)\Big] \end{equation}	\begin{equation} s > 0 \end{equation}

Table 338. L\(_1\)

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L\(_1\)


\begin{equation*}
1
\end{equation*}


\begin{equation*}
\frac{1}{s}
\end{equation*}


\begin{equation*}
s > 0
\end{equation*}

Table 339. L\(_2\)

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L\(_2\)


\begin{equation*}
e^{at}
\end{equation*}


\begin{equation*}
\frac{1}{s - a}
\end{equation*}


\begin{equation*}
s > a
\end{equation*}

Table 340. L\(_3\)

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L\(_3\)


\begin{equation*}
t^n
\end{equation*}


\begin{equation*}
\frac{n!}{s^{n+1}}
\end{equation*}


\begin{equation*}
s > 0
\end{equation*}

Table 341. L\(_4\)

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L\(_4\)


\begin{equation*}
\sin(bt)
\end{equation*}


\begin{equation*}
\frac{b}{s^2+b^2}
\end{equation*}


\begin{equation*}
s > 0
\end{equation*}

Table 342. L\(_5\)

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L\(_5\)


\begin{equation*}
\cos(bt)
\end{equation*}


\begin{equation*}
\frac{s}{s^2+b^2}
\end{equation*}


\begin{equation*}
s > 0
\end{equation*}

Table 343. L\(_6\)

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L\(_6\)


\begin{equation*}
e^{at}\ t^n
\end{equation*}


\begin{equation*}
\frac{n!}{(s-a)^{n+1}}
\end{equation*}


\begin{equation*}
s > a
\end{equation*}

Table 344. L\(_7\)

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L\(_7\)


\begin{equation*}
e^{at}\sin(bt)
\end{equation*}


\begin{equation*}
\frac{b}{(s-a)^2+b^2}
\end{equation*}


\begin{equation*}
s > a
\end{equation*}

Table 345. L\(_8\)

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L\(_8\)


\begin{equation*}
e^{at}\cos(bt)
\end{equation*}


\begin{equation*}
\frac{s-a}{(s-a)^2+b^2}
\end{equation*}


\begin{equation*}
s > a
\end{equation*}

Table 346. L\(_1\)

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	Function (\(t\)-Domain) \begin{equation} f(t) \end{equation}	Laplace Transform (\(s\)-Domain) \begin{equation} \lap{f(t)} = F(s) \end{equation}	Existence Condition
L\(_1\)	\begin{equation} 1 \end{equation}	\begin{equation} \frac{1}{s} \end{equation}	\begin{equation} s > 0 \end{equation}

Table 347. L\(_2\)

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	Function (\(t\)-Domain) \begin{equation} f(t) \end{equation}	Laplace Transform (\(s\)-Domain) \begin{equation} \lap{f(t)} = F(s) \end{equation}	Existence Condition
L\(_2\)	\begin{equation} e^{at} \end{equation}	\begin{equation} \frac{1}{s - a} \end{equation}	\begin{equation} s > a \end{equation}

Table 348. L\(_3\)

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	Function (\(t\)-Domain) \begin{equation} f(t) \end{equation}	Laplace Transform (\(s\)-Domain) \begin{equation} \lap{f(t)} = F(s) \end{equation}	Existence Condition
L\(_3\)	\begin{equation} t^n \end{equation}	\begin{equation} \frac{n!}{s^{n+1}} \end{equation}	\begin{equation} s > 0 \end{equation}

Table 349. L\(_4\)

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	Function (\(t\)-Domain) \begin{equation} f(t) \end{equation}	Laplace Transform (\(s\)-Domain) \begin{equation} \lap{f(t)} = F(s) \end{equation}	Existence Condition
L\(_4\)	\begin{equation} \sin(bt) \end{equation}	\begin{equation} \frac{b}{s^2+b^2} \end{equation}	\begin{equation} s > 0 \end{equation}

Table 350. L\(_5\)

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	Function (\(t\)-Domain) \begin{equation} f(t) \end{equation}	Laplace Transform (\(s\)-Domain) \begin{equation} \lap{f(t)} = F(s) \end{equation}	Existence Condition
L\(_5\)	\begin{equation} \cos(bt) \end{equation}	\begin{equation} \frac{s}{s^2+b^2} \end{equation}	\begin{equation} s > 0 \end{equation}

Table 351. L\(_6\)

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	Function (\(t\)-Domain) \begin{equation} f(t) \end{equation}	Laplace Transform (\(s\)-Domain) \begin{equation} \lap{f(t)} = F(s) \end{equation}	Existence Condition
L\(_6\)	\begin{equation} e^{at}\ t^n \end{equation}	\begin{equation} \frac{n!}{(s-a)^{n+1}} \end{equation}	\begin{equation} s > a \end{equation}

Table 352. L\(_7\)

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	Function (\(t\)-Domain) \begin{equation} f(t) \end{equation}	Laplace Transform (\(s\)-Domain) \begin{equation} \lap{f(t)} = F(s) \end{equation}	Existence Condition
L\(_7\)	\begin{equation} e^{at}\sin(bt) \end{equation}	\begin{equation} \frac{b}{(s-a)^2+b^2} \end{equation}	\begin{equation} s > a \end{equation}

Table 353. L\(_8\)

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	Function (\(t\)-Domain) \begin{equation} f(t) \end{equation}	Laplace Transform (\(s\)-Domain) \begin{equation} \lap{f(t)} = F(s) \end{equation}	Existence Condition
L\(_8\)	\begin{equation} e^{at}\cos(bt) \end{equation}	\begin{equation} \frac{s-a}{(s-a)^2+b^2} \end{equation}	\begin{equation} s > a \end{equation}

Table 354. R\(_1\)

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	Function (\(t\)-Domain)	Laplace Transform (\(s\)-Domain)	Existence
	\begin{equation} f(t) \end{equation}	\begin{equation} \lap{f(t)} = F(s) \end{equation}	Condition
R\(_{1}\)	\begin{equation} f'(t) \end{equation}	\begin{equation} sF(s) - f(0) \end{equation}	\begin{equation} s > 0 \end{equation}

Table 355. R\(_1\)

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	Function (\(t\)-Domain)	Laplace Transform (\(s\)-Domain)	Existence
	\begin{equation} f(t) \end{equation}	\begin{equation} \lap{f(t)} = F(s) \end{equation}	Condition
R\(_{2}\)	\begin{equation} f''(t) \end{equation}	\begin{equation} s^2F(s) - sf(0) - f'(0) \end{equation}	\begin{equation} s > 0 \end{equation}

Table 356. R\(_1\)

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	Function (\(t\)-Domain)	Laplace Transform (\(s\)-Domain)	Existence
	\begin{equation} f(t) \end{equation}	\begin{equation} \lap{f(t)} = F(s) \end{equation}	Condition
R\(_{3}\)	\begin{equation} f'''(t) \end{equation}	\begin{equation} s^3F(s) - s^2f(0) - sf'(0) - f''(0) \end{equation}	\begin{equation} s > 0 \end{equation}

Table 357. R\(_1\)

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	Function (\(t\)-Domain)	Laplace Transform (\(s\)-Domain)	Existence
	\begin{equation} f(t) \end{equation}	\begin{equation} \lap{f(t)} = F(s) \end{equation}	Condition
R\(_{4}\)	\begin{equation} e^{at} f(t) \end{equation}	\begin{equation} F(s-a) \end{equation}	\begin{equation} s > 0 \end{equation}

Table 358. R\(_1\)

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	Function (\(t\)-Domain)	Laplace Transform (\(s\)-Domain)	Existence
	\begin{equation} f(t) \end{equation}	\begin{equation} \lap{f(t)} = F(s) \end{equation}	Condition
R\(_{5}\)	\begin{equation} t^n f(t) \end{equation}	\begin{equation} (-1)^n \frac{d^n}{ds^n}\Big[F(s)\Big] \end{equation}	\begin{equation} s > 0 \end{equation}

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Subsubsection Laplace Transforms

Worksheet Key Terms & Concepts

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✳️ Summary of the Key Ideas.

Common Forms: A table of common Laplace transforms is provided, which doubles as a reference for inverse transforms. The focus is on recognizing forms that match the table entries for functions like \(\sin(bt), \cos(bt)\text{,}\) and others.

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Direct Computation: When the function of \(s\) directly matches a form in the common Laplace transform table, the inverse Laplace transform can be easily computed.

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Modifying Functions: When a function doesn’t match a known form, minor modifications, such as multiplying by missing constants or splitting fractions, can help.

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Completing the Square: When dealing with quadratic expressions in the denominator, especially when the discriminant is negative, completing the square can transform the expression into a form that matches known inverse Laplace transforms. Several examples demonstrate this technique.

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Partial Fraction Decomposition: For more complex rational functions, partial fraction decomposition breaks down the function into simpler fractions that match the common transform forms.

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Table 359. Matching Guide: \(s\)-function \(\rightarrow\) Inverse Transform

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Denominator Form	Example	Transform
\begin{equation} \frac{c}{s} \end{equation}	\begin{equation} \frac{5}{s} \end{equation}	L\(_1\)
\begin{equation} \frac{c}{s^P} \end{equation}	\begin{equation} \frac{-10}{s^5} \end{equation}	L\(_3\)
\begin{equation} \frac{c}{s\pm a} \end{equation}	\begin{equation} \frac{1}{s + 1.8} \end{equation}	L\(_2\)
\begin{equation} \frac{c}{(s\pm a)^P} \end{equation}	\begin{equation} \frac{6.77}{(s - 3)^9} \end{equation}	L\(_6\)
\begin{equation} \frac{cb}{s^2 + b^2} \end{equation}	\begin{equation} \frac{\pi}{s^2 + 4} \end{equation}	L\(_4\)
\begin{equation} \frac{cs}{s^2 + b^2} \end{equation}	\begin{equation} \frac{6s}{s^2 + 3} \end{equation}	L\(_5\)
\begin{equation} \frac{cb}{(s \pm a)^2 + b^2} \end{equation}	\begin{equation} \frac{1}{(s - 3)^2 + 1} \end{equation}	L\(_7\)
\begin{equation} \frac{c(s \pm a)}{(s \pm a)^2 + b^2} \end{equation}	\begin{equation} \frac{-0.33(s + 17)}{(s + 17)^2 + 12} \end{equation}	L\(_8\)

Tips for Preparing the Backward Transform.

Completing the square is an essential technique for transforming quadratic expressions that don’t directly match a form in the table of common Laplace transforms. However, it’s not the only strategy available. In this section, we’ll explore another important technique: partial fraction decomposition. This method is useful for breaking down complex fractions into simpler components that can each be matched with forms in the Laplace transform table.

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Two other forms we may wish to match when we study Laplace transforms are

\begin{equation*} \frac{b}{(s-a)^2 + b^2} \mbox{ and } \frac{s-a}{(s-a)^2 + b^2}. \end{equation*}

As before, we work toward making the denominator match first, and then we sort out the numerator second.

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When we want to take the inverse Laplace transform of a rational function with a second-degree polynomial in the denominator, we may complete the square or we may do a partial fraction decomposition. How will we know which is appropriate? Here are a few guidelines for you to consider.

Does the denominator factor in an obvious way? If so, factor the denominator and do a partial fraction decomposition if necessary.

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If the denominator does not factor in an obvious way, try completing the square.

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If you end up with addition outside of the parentheses, as in \((s - a)^2 + b^2,\) then you should aim to match \(L7\) and/or \(L8\).

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If instead you end up with subtraction outside the parentheses, as in \((s - a)^2 - b^2,\) then you should factor and do a partial fraction decomposition. You may consider using the quadratic formula if the factorization is not obvious to you.

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If you end up with no terms outside the parentheses, as in \((s - a)^2,\) then use \(L6\).

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Subsubsection Laplace Transforms

✳️ Summary of the Key Ideas.

PIECEWISE AND UNIT STEP STUFF

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Subsubsection First-Order Linear Systems

Worksheet Key Terms & Concepts

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You have attempted of activities on this page.

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Denominator Form	Example	Transform
\begin{equation} \frac{c}{s} \end{equation}	\begin{equation} \frac{5}{s} \end{equation}	L\(_1\)
\begin{equation} \frac{c}{s^P} \end{equation}	\begin{equation} \frac{-10}{s^5} \end{equation}	L\(_3\)
\begin{equation} \frac{c}{s\pm a} \end{equation}	\begin{equation} \frac{1}{s + 1.8} \end{equation}	L\(_2\)
\begin{equation} \frac{c}{(s\pm a)^P} \end{equation}	\begin{equation} \frac{6.77}{(s - 3)^9} \end{equation}	L\(_6\)
\begin{equation} \frac{cb}{s^2 + b^2} \end{equation}	\begin{equation} \frac{\pi}{s^2 + 4} \end{equation}	L\(_4\)
\begin{equation} \frac{cs}{s^2 + b^2} \end{equation}	\begin{equation} \frac{6s}{s^2 + 3} \end{equation}	L\(_5\)
\begin{equation} \frac{cb}{(s \pm a)^2 + b^2} \end{equation}	\begin{equation} \frac{1}{(s - 3)^2 + 1} \end{equation}	L\(_7\)
\begin{equation} \frac{c(s \pm a)}{(s \pm a)^2 + b^2} \end{equation}	\begin{equation} \frac{-0.33(s + 17)}{(s + 17)^2 + 12} \end{equation}	L\(_8\)

Section Quick References

📝

Subsubsection Basics

Worksheet Key Terms & Concepts

✳️ Differential Equations and their Components.

✳️ Order & Linearity.

Subsubsection Solutions to Differential Equations

Worksheet Key Terms & Concepts

✳️ Solutions & Initial Conditions.

✳️ Direct Integration.

Subsubsection Separation of Variables

Worksheet Key Terms & Concepts

✳️ First-Order Differential Equations.

✳️ Separation of Variables.

Subsubsection Integrating Factor

Worksheet Key Terms & Concepts

✳️ Integrating Factor.

Subsubsection Qualitative Methods

Worksheet Key Terms & Concepts

Subsubsection Numerical Methods

Worksheet Key Terms & Concepts

Subsubsection Linear Homogeneous Differential Equations with Constant Coefficients

Worksheet Key Terms & Concepts

✳️ Summary of the Key Ideas.

✍🏻 Method 7. LHCC Method.

✳️ Properties of LHCC Solutions.

✳️ Key Takeaways.

Subsubsection Undermined Coefficients

Worksheet Key Terms & Concepts

✳️ Summary of the Key Ideas.

Subsection LNCC Equations

Subsubsection Laplace Transforms

Worksheet Key Terms & Concepts

✳️ Summary of the Key Ideas.

Common Laplace Transforms.

L\(_1\)\(\ds\quad\lap{ 1 } = \frac{1}{s}, \quad s > 0\).

L\(_2\)\(\ds\quad\lap{ e^{at} } = \frac{1}{s - a}, \quad s >a \).

L\(_3\)\(\ds\quad\lap{ t^n } = \frac{n!}{s^{n+1}}, \quad s >0, \quad n = 1, 2, 3, \ldots \).

L\(_4\)\(\ds\quad\lap{ \sin(bt) } = \frac{b}{s^2+b^2}, \quad s >0\).

L\(_5\)\(\ds\quad\lap{ \cos(bt) } = \frac{s}{s^2+b^2}, \quad s >0\).

L\(_6\)\(\ds\quad\lap{ e^{at}\ t^n } = \frac{n!}{(s-a)^{n+1}}, \quad s >a, \quad n = 0, 1, 2, 3, \ldots \).

L\(_7\)\(\ds\quad\lap{ e^{at}\sin(bt) } = \frac{b}{(s-a)^2+b^2}, \quad s >a\).

L\(_8\)\(\ds\quad\lap{ e^{at}\cos(bt) } = \frac{s-a}{(s-a)^2+b^2}, \quad s >a\).

Laplace Transform of Derivatives.

L\(_9\)\(\ds\quad\lap{ f'(t) } = sF(s) - f(0), \quad s > 0\).

L\(_{10}\)\(\ds\quad\lap{ f''(t) } = s^2F(s) - sf(0) - f'(0), \quad s > 0\).

L\(_{11}\)\(\ds\quad\lap{ f'''(t) } = s^3F(s) - s^2f(0) - sf'(0) - f''(0), \quad s > 0\).

L\(_{12}\)\(\ds\quad\lap{ t^n f(t) } = (-1)^n \frac{d^n}{ds^n}F(s), \quad s > 0, \quad n = 0, 1, 2, 3, \ldots \).

Subsubsection Laplace Transforms

Worksheet Key Terms & Concepts

✳️ Summary of the Key Ideas.

Tips for Preparing the Backward Transform.

Subsubsection Laplace Transforms

✳️ Summary of the Key Ideas.

Subsubsection First-Order Linear Systems

Worksheet Key Terms & Concepts

Worksheet Key Terms & Concepts

Worksheet Key Terms & Concepts

Worksheet Key Terms & Concepts

Worksheet Key Terms & Concepts

Worksheet Key Terms & Concepts

Worksheet Key Terms & Concepts

Worksheet Key Terms & Concepts

Worksheet Key Terms & Concepts

Worksheet Key Terms & Concepts

Worksheet Key Terms & Concepts

Worksheet Key Terms & Concepts