DE-BOOK The LNCC Equation

Section 10.1 The LNCC Equation

You already know that a linear homogeneous constant coefficient (LHCC) equation has the form:

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

All the terms are on the left, and the right-hand side is zero. This equation is called homogeneous because the solution only includes terms that cancel each other out.

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A linear nonhomogeneous constant coefficient (LNCC) equation looks almost the same, except the right-hand side is now a function:

\begin{equation*} a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y' + a_0 y = f(x) \end{equation*}

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The function \(f(x)\) is called the forcing function because it introduces an external input that “forces” the system to respond in a particular way.

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Subsection Guessing a Solution that Accounts for the Forcing Function

Checkpoint 146. 📖❓ Homogeneous vs Nonhomogeneous.

Which structure or property of a differential equation separates homogeneous from nonhomogeneous equations?

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The presence of a non-zero forcing function.
Correct! A non-zero free term on the right-hand side of the equation indicates a nonhomogeneous equation.
The dependent variable.
Incorrect. The dependent variable is present in both homogeneous and nonhomogeneous equations.
The order of the equation.
Incorrect. Both homogeneous and nonhomogeneous equations can have the same order.
The linearity of the equation.
Incorrect. Both homogeneous and nonhomogeneous equations can be linear.

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Note: For brevity, we will refer to the left side of equation (10.1) as 🍄. Also, when we say “plug \(y\) into 🍄”, we mean calculate and substitute \(y''\text{,}\) \(y'\text{,}\) \(y\) into \(y'' - 4y' + 3y\text{,}\) then simplify.

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Imagine you had to make your best guess as to which function, \(y\text{,}\) satisfies the equation

\begin{equation} \ub{y'' - 4y' + 3y}_{\Large 🍄} = 9x + 6.\tag{10.1} \end{equation}

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Remember, after differentiating your guess and plugging it into 🍄, all terms must simplify to \(9x + 6\text{.}\) That is,

\begin{equation*} (\text{guess})'' - 4(\text{guess})' + 3(\text{guess})\quad\rightarrow\quad\text{simplifies to}\quad 9x + 6 \end{equation*}

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Logically, for the sum of three terms to simplify to \(9x + 6\text{,}\) they must share a similar structure as \(9x + 6\) (i.e., a degree \(1\) polynomial).

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or contain terms that completely cancel out, but we will ignore these for now.

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Checkpoint 147. 📖❓ Select the most appropriate solution.

Based on this logic, which function has the most appropriate form for a particular solution to the LNCC equation?

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\begin{equation*} y'' - 2y' + y = x^3 + x\text{.} \end{equation*}

Hint: Think about the type of functions you need to add together to get \(x^3 + x\text{.}\)

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\(\quad e^{3x}\)
Incorrect. This function does not resemble the forcing function.
\(\quad 3x^2 + 1 \)
Incorrect. Plugging \(3x^2 + 1\) into \(y'' - 2y' + y\) will not give you an \(x^3\) term you need to match the forcing function, \(x^3 + x\text{.}\)
\(\quad x^3 + 6x^2 + 19x + 26\)
Correct! This 3rd degree polynomial produces the forcing function, \(x^3 + x\text{,}\) when substituted into the left-hand side.
\(\quad x^4 + x^2\)
Incorrect. Plugging \(x^4 + x^2\) into \(y'' - 2y' + y\) will leave you with an \(x^4\) term that does not exist in the forcing function, \(x^3 + x\text{.}\)

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In this context, a “similar function-type” to \(9x + 6\) means our guess should also be a degree \(1\) polynomial. Since we don’t know which one, we use the general degree \(1\) form:

\begin{equation*} y_p = Ax + B \end{equation*}

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Plugging this \(y_p\) into 🍄 and taking some derivatives, we get

\begin{align*} y'' - 4y' + 3y \amp = (Ax + B)'' - 4(Ax + B)' + 3(Ax + B) \\ \amp = 0 - 4A + 3Ax + 3B \\ \amp = 3Ax + 3B - 4A \end{align*}

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This is not \(9x + 6\text{,}\) but maybe we can find \(A\) and \(B\) values that get us there. Setting the expression equal to \(9x + 6\text{,}\) we have

\begin{equation*} {\DLO 3A}x + {\DLBa 3B - 4A} = {\DLO 9}x + {\DLBa 6} \end{equation*}

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Matching the \({\DLO x\ \text{coefficients}}\) and \({\DLBa \text{constants}}\text{,}\) gives us the system:

\begin{align*} 3A \amp = 9 \quad \Rightarrow \quad A = 3 \\ -4A + 3B \amp = 6 \quad \Rightarrow \quad -12 + 3B = 6 \Rightarrow B = 6 \end{align*}

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Substituting \(A=3\) and \(B=6\) into \(y_p\) completes the solution:

\begin{equation*} y_p = 3x + 6 \end{equation*}

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Subsection Confirming our Guess

Let’s double-check that the function, \(y_p\text{,}\) we found really satisfies the equation.

\begin{align*} (3x+6)'' - 4(3x+6)' + 3(3x+6) \amp = 9x + 6\\ 0 - 12 + 9x + 18 \amp = 9x + 6\\ {\color{RoyalBlue}\text{all terms simplify to }\ \rightarrow}\quad 9x + 6 \amp = 9x + 6\quad {\color{RoyalBlue}\leftarrow\ \text{forcing function}} \end{align*}

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This confirms \(y_p\) as a solution. In fact, it is the most important solution because it accounts for the specific forcing function \(9x + 6\) and must be included in every solution to equation (10.1). Such solutions are called particular solutions, but they are only part of the whole story.

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Subsection The Hidden Parts of Nonhomogeneous Solutions

We established that the particular solution, \(y_p = 3x + 6\text{,}\) is what you plug into 🍄 to get \(9x + 6\text{.}\) However, there could be additional terms you can plug into 🍄 that cancel out after taking derivatives and simplifying. Specifically,

\begin{equation*} y = \ub{3x + 6}_{\large y_p} + \left[ \begin{array}{c} \text{terms that simplify to}\ 0 \\ \text{when plugged into 🍄} \end{array} \right] \end{equation*}

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We already know how to find these functions since they are just the solutions to the homogeneous version of the same equation:

\begin{equation*} y'' - 4y' + 3y = 0 \end{equation*}

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This LHCC equation is solved using its characteristic equation:

\begin{align*} r^2 - 4r + 3 = 0 \amp\quad\Rightarrow\quad (r - 1)(r - 3) = 0 \\ \amp\quad\Rightarrow\quad r = 1,\ 3 \end{align*}

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This homogeneous solution, we’ll call \(y_h\text{,}\) is given by

\begin{equation*} y_h = c_1 e^{x} + c_2 e^{3x} \end{equation*}

and is the other half of the LNCC solution. It provides the terms that cancel out when plugged into 🍄 and have no impact on the forcing function.

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Subsection Combining the Parts Into a General Solution

To summarize, the general solution of the equation

\begin{equation*} y'' - 4y' + 3y = 9x + 6 \end{equation*}

consists of two parts:

The homogeneous part: \(\quad y_h = c_1 e^{x} + c_2 e^{3x}\quad\) (canceling terms), and

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The particular part: \(\quad y_p = 3x + 6\quad\) (account for \(9x + 6\))

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and we write it as the sum of these parts:

\begin{equation*} y = y_h + y_p = c_1 e^{x} + c_2 e^{3x} + 3x + 6. \end{equation*}

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Finally, let’s see how the different parts of the general solution behave when plugged into the left side of the equation. First, let’s plug in \(y_h + y_p\)

\begin{equation*} (y_h + y_p)'' - 4(y_h + y_p)' + 3(y_h + y_p) = 9x + 6 \end{equation*}

and regroup the \(y_h\) and \(y_p\) terms:

\begin{equation*} \ub{y_h'' - 4y_h' + 3y_h}_{\ds\text{1️⃣}} + \ub{y_p'' - 4y_p' + 3y_p}_{\ds\text{2️⃣}} = 9x + 6 \end{equation*}

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Now, we will show that the homogeneous terms, 1️⃣, simplify to zero:

\begin{align*} \text{1️⃣}:\ y_h'' - 4y_h' + 3y_h \amp = \left(c_1e^{x} + c_2e^{3x}\right)'' - 4\left(c_1e^{x} + c_2e^{3x}\right)' + 3\left(c_1e^{x} + c_2e^{3x}\right) \\ \amp = c_1e^{x} + 9c_2e^{3x} - 4\left(c_1e^{x} + 3c_2e^{3x}\right) + 3\left(c_1e^{x} + c_2e^{3x}\right) \\ \amp = c_1e^{x} + 9c_2e^{3x} - 4c_1e^{x} - 12c_2e^{3x} + 3c_1e^{x} + 3c_2e^{3x} \\ \amp = 0 \quad \text{✅} \end{align*}

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Next, we will show that the particular terms, 2️⃣, simplify to \(9x + 6\text{:}\)

\begin{align*} \text{2️⃣}:\ y_p'' - 4y_p' + 3y_p \amp = (3x + 6)'' - 4(3x + 6)' + 3(3x + 6) \\ \amp = 0 - 12 + 9x + 18 \\ \amp = 9x + 6 \quad \text{✅} \end{align*}

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So adding the results of 1️⃣ and 2️⃣ leaves us with the forcing function, \(9x + 6\text{.}\)

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Subsection The General Solution to LNCC Equations

Let’s summarize what we’ve learned so far:

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✳️ LNCC General Solution Structure.

A linear nonhomogeneous constant coefficient (LNCC) equation has the form:

\begin{equation} a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_0 y = f(x)\tag{10.2} \end{equation}

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Its general solution is built from two parts:

\begin{equation} y = y_h + y_p\tag{10.3} \end{equation}

\(y_h\) is the solution to the corresponding homogeneous (LHCC) equation:

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

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\(y_p\) is the particular solution that produces the forcing function \(f(x)\text{.}\)

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Checkpoint 148. 📖❓ Constructing the General Solution.

Suppose the homogeneous and particular solutions of an LNCC equation are given by

\begin{equation*} y_h = c_1e^{-x} + c_2e^{2x} \quad \text{and} \quad y_p = 5x - 3\text{.} \end{equation*}

Select the general solution for such an equation.

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\(\quad y = 5x - 3\text{.}\)
Incorrect. This is only a particular solution, not a complete general solution.
\(\quad y = c_1e^{-x} + c_2x e^{2x} + 5x - 3\text{.}\)
Incorrect. One of the terms in this expression is incorrect.
\(\quad y = c_1e^{-x} + c_2e^{2x} + 5x - 3\text{.}\)
Correct! The general solution is the sum of the homogeneous and particular solutions.
\(\quad y = c_1e^{-x} + c_2e^{2x} + c_3(5x - 3)\text{.}\)
Incorrect. You do not multiply the particular solution by a constant.

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Here’s a quick example to help solidify this concept.

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🌌 Example 149. Finding the general solution, given \(y_p\).

Find the general solution to the differential equation

\begin{equation*} y'' - 4y' - 12y = 3{e^{5t}} \end{equation*}

given that the particular solution is known to be \(y_p(t) = -\frac{3}{7}{{e}^{5t}}\text{.}\)

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Solution.

The general solution has the form:

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

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Since \(y_p\) is given, we only need to solve the homogeneous equation:

\begin{equation*} y'' - 4y' - 12y = 0\text{.} \end{equation*}

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Using the characteristic equation:

\begin{align*} r^2 - 4r - 12 \amp = 0 \\ (r - 6)(r + 2) \amp = 0 \quad \Rightarrow \quad r_1 = 6, r_2 = -2 \end{align*}

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The homogeneous solution is:

\begin{equation*} y_h(t) = c_1e^{-2t} + c_2e^{6t}\text{.} \end{equation*}

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Therefore, the general solution is:

\begin{equation*} y(t) = \ub{c_1e^{-2t} + c_2e^{6t}\vphantom{\frac{3}{7}}}_{\Large y_h}\ \ub{-\frac{3}{7}{e^{5t}}}_{\Large y_p} \end{equation*}

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While the structure of LNCC solutions is now clear, actually determining \(y_p\) requires a systematic approach. The next section introduces the Method of Undetermined Coefficients, a technique for efficiently constructing \(y_p\) when \(f(x)\) has a specific form.

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Checkpoint 150. 🤔💭 Reading Questions.

(a) 🤔💭 Role of the Particular Solution.

Which statement best describes the role of the particular solution to an LNCC equation?

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Part of the general solution that accounts for the forcing function.
Correct! The particular solution is chosen to match the form of \(f(x)\) and account for its influence in the equation.
General solution of the LNCC equation.
Incorrect. The general solution of the homogeneous equation is called the complementary solution, not the particular solution.
Roots of the characteristic equation.
Incorrect. The characteristic equation is related to the complementary solution and does not involve the particular solution.
Coefficients of the differential equation.
Incorrect. The particular solution addresses the specific form of \(f(x)\) and is part of solving the non-homogeneous equation, but its purpose isn’t simplification.

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(b) 🤔💭 Particular Solution Properties.

Which statements are true about the particular solution, \(y_p\text{,}\) of

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

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It is part of the general solution.
Correct. \(y_p\) is part of \(y_h+y_p\text{.}\)
All of its terms simplify to \(0\) when plugged into the equation.
Incorrect. That property describes \(y_h\text{.}\)
It is constructed from the roots of a characteristic equation.
Incorrect. Roots/characteristic equation come from \(y_h\text{.}\)
It mimics the structure of the forcing function, \(f(x)\text{.}\)
Correct. \(y_p\) is chosen to match \(f(x)\text{.}\)
It contains constants of integration (e.g., \(c_1\text{,}\) \(c_2\text{,}\) etc.).
Incorrect. \(y_p\) has no arbitrary constants.

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(c) 🤔💭 Homogeneous Solution Properties.

Which statements are true about the homogeneous solution, \(y_h\text{,}\) of

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

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It is part of the general solution.
Correct. \(y_h\) is part of \(y_h+y_p\text{.}\)
All of its terms simplify to \(0\) when plugged into the equation.
Correct. \(y_h\) makes the left side simplify to \(0\text{.}\)
It is constructed from the roots of a characteristic equation.
Correct. \(y_h\) comes from the characteristic equation.
It mimics the structure of the forcing function, \(f(x)\text{.}\)
Incorrect. Mimicking \(f(x)\) is for \(y_p\text{.}\)
It contains constants of integration (e.g., \(c_1\text{,}\) \(c_2\text{,}\) etc.).
Correct. \(y_h\) includes constants like \(c_1\text{,}\) \(c_2\text{.}\)

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