DE-BOOK The LNCC Equation

Section 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:

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

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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Checkpoint 154. 📖❓ 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.
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Correct! A non-zero free term on the right-hand side of the equation indicates a nonhomogeneous equation.
The dependent variable.
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Incorrect. The dependent variable is present in both homogeneous and nonhomogeneous equations.
The order of the equation.
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Incorrect. Both homogeneous and nonhomogeneous equations can have the same order.
The linearity of the equation.
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Incorrect. Both homogeneous and nonhomogeneous equations can be linear.

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Subsection Guessing a Solution that Accounts for the Forcing Function

Note: For brevity, we will refer to the left side of equation (39) 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

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\begin{equation} \ub{y'' - 4y' + 3y}_{\Large 🍄} = 9x + 6\text{.}\tag{39} \end{equation}

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

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

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 155. 📖❓ Select the most appropriate solution.

Based on this logic, which function is the only possible candidate for the 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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\(\ds\quad e^{3x}\)
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Incorrect. This function does not resemble the forcing function.
\(\ds\quad 3x^2 + 1 \)
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Incorrect. Plugging \(3x^2 + 1\) into \(y'' - 2y' + y\) will not give you an \(x^3\) term you need to match the forcing function, \(x^3 + x\text{.}\)
\(\ds\quad x^3 + 6x^2 + 18x + 24\)
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Correct! This 3rd degree polynomial has the same polynomial structure as the forcing function, \(x^3 + x\text{.}\)
\(\ds\quad x^4 + x^2\)
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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

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\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*}

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

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

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

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\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*}

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.

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\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*}

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 (39). 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,

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\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*}

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

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

This LHCC equation is solved using its characteristic equation:

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\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*}

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

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

and is the other half of LNCC solution. They provide the terms that cancel out when plugged into 🍄 and have no impact on the forcing function at all.

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Subsection Combining the Parts Into a General Solution

To summarize, the general solution of 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:

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

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 \(y_h + y_p\)

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

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

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

Now, we will now 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*}

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

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\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*}

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:

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\begin{equation} a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_0 y = f(x)\tag{40} \end{equation}

Its general solution is built from two parts:

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\begin{equation} y = y_h + y_p\tag{41} \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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\(y_p\) the particular solution that produces the forcing function \(f(x)\text{.}\)

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Checkpoint 156. 📖❓ Contructing 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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\(\ds\quad y = 5x - 3\text{.}\)
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Incorrect. This is only the particular solution, not the complete general solution.
\(\ds\quad y = c_1e^{-x} + c_2x e^{2x} + 5x - 3\text{.}\)
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Incorrect. One of the terms in this expression is incorrect.
\(\ds\quad y = c_1e^{-x} + c_2e^{2x} + 5x - 3\text{.}\)
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Correct! The general solution is the sum of the homogeneous and particular solutions.
\(\ds\quad y = c_1e^{-x} + c_2e^{2x} + c_3(5x - 3)\text{.}\)
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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 157. 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:

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

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

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

Using the characteristic equation:

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\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*}

The homogeneous solution is:

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

Therefore, the general solution is:

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\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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Subsection 📤 Wrap-Up

🗝️ \(\textbf{Key Takeaways...}\)

A linear differential equation is nonhomogeneous if the right-hand side is not zero—it includes a forcing function, \(f(x)\text{.}\)

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You can often guess a particular solution that matches the structural forms of the forcing function.

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The general solution is \(y = y_h + y_p\text{.}\)

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Check Your Understanding.

Checkpoint 158. 🤔💭 Separation of Variables 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.
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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.
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Incorrect. The general solution of the homogeneous equation is called the complementary solution, not the particular solution.
Roots of the characteristic equation.
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Incorrect. The characteristic equation is related to the complementary solution and does not involve the particular solution.
Coefficients of the differential equation.
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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.
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selected.
All of its terms simplify to \(0\) when plugged into the equation.
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selected.
It is constructed from the roots of a characteristic equation.
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selected.
It mimics the structure of the forcing function, \(f(x)\text{.}\)
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selected.
It contains constants of integration (e.g., \(c_1\text{,}\) \(c_2\text{,}\) etc.).
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selected.

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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.
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selected.
All of its terms simplify to \(0\) when plugged into the equation.
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selected.
It is constructed from the roots of a characteristic equation.
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selected.
It mimics the structure of the forcing function, \(f(x)\text{.}\)
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selected.
It contains constants of integration (e.g., \(c_1\text{,}\) \(c_2\text{,}\) etc.).
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selected.

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