Section 10.2 Selecting the Particular Solution Form

In the previous section, we saw that the general solution to a linear nonhomogeneous constant coefficient (LNCC) equation has two parts: the homogeneous solution \(y_h\) and the particular solution \(y_p\text{.}\) The homogeneous solution is found using the characteristic equation, while the particular solution is constructed by (1) selecting a form similar to the forcing function, \(f(x)\text{,}\) and (2) determining the coefficients of that form.

In this section, we focus on selecting a form for \(y_p\text{.}\) The key idea is to match the structural form of the forcing function \(f(x)\text{,}\) so that when it is substituted into the differential equation, all the resulting terms simplify to \(f(x)\text{.}\)

Subsection Selecting Basic Forms

The first step in choosing \(y_p\) is recognizing the structure of \(f(x)\text{.}\) Since \(y_p\) and its derivatives must eventually simplify to match \(f(x)\text{,}\) \(y_p\) should have a similar form. If \(f(x)\) is a polynomial, exponential, sine, or cosine function, then \(y_p\) should be too.

Table 151 provides a guide for selecting an initial form of \(y_p\) based on the type of \(f(x)\text{.}\) In each case, \(A, B, C, \dots\) represent the undetermined coefficients we’ll solve for later.

Table 151. \(y_p\) forms based on \(f(x)\text{,}\) where \(a, b, c, d, \alpha, \beta\) are known constants. \(A, B, C, D\) must be determined. Degree 4+ polynomials follow a similar pattern.

	Forcing Function Type,\(f(x)\)	Particular Solution Form,\(y_p\)
1	\(a\) (constant)	\(A\)
2	\(ax + b\)	\(Ax + B\)
3	\(ax^2 + bx + c\)	\(Ax^2 + Bx + C\)
4	\(ax^3 + bx^2 + cx + d\)	\(Ax^3 + Bx^2 + Cx + D\)
5	\(a e^{\ds\alpha x}\)	\(A e^{\ds\alpha x}\)
6	\(a \sin(\beta x) + b \cos(\beta x)\)	\(A \sin(\beta x) + B \cos(\beta x)\)

One important note: even if a term is missing in \(f(x)\text{,}\) you should still include it in \(y_p\) if it’s part of the general form. For example:

🔸

If \(f(x) = x^2 + 1\text{,}\) the \(x\) term is missing, but \(y_p\) must still include it:

\begin{equation*} y_p = Ax^2 + Bx + C \text{.} \end{equation*}

🔸

If \(f(x) = \cos(3x)\text{,}\) then \(y_p = A\cos(3x) + B\sin(3x)\text{.}\) Even though \(\sin(3x)\) isn’t in \(f(x)\text{,}\) it’s needed because its derivative produces cosine terms.

🌌 Example 152. Selecting Basic Forms of \(y_p\).

Use Table 151 to determine the appropriate form of \(y_p\) in each case.

\(\ds y'' + y' + 5y = 3e^{6x}\)

Solution.

Since \(f(x)\) is exponential, we use row 5:

\begin{equation*} y_p = Ae^{6x} \text{.} \end{equation*}

\(\ds y'' + 3y' - y = 2\sin(2x)\)

Solution.

This matches row 6. Even though \(\cos(2x)\) doesn’t appear in \(f(x)\text{,}\) we include it:

\begin{equation*} y_p = A\sin(2x) + B\cos(2x) \text{.} \end{equation*}

\(\ds \omega''' + \omega' = 5\theta^2 + \theta + 1\)

Solution.

\(f(x)\) is a degree 2 polynomial, so from row 3:

\begin{equation*} \omega_p = A\theta^2 + B\theta + C \text{.} \end{equation*}

\(\ds 2\frac{dy}{dx} + 5y + 3 = 0\)

Solution.

Move the constant to the right side: \(2y' + 5y = -3\text{.}\) Now \(f(x)\) is constant (row 1):

\begin{equation*} y_p = A \text{.} \end{equation*}

\(\ds y'' -3y' + 0.5y = 14x^4\)

Solution.

4th degree polynomials are not represented in the table, but follow a similar pattern:

\begin{equation*} y_p = Ax^4 + Bx^3 + Cx^2 + Dx + E \text{.} \end{equation*}

Checkpoint 153. 📖❓ Matching single term forcing functions to the particular form.

Drag each forcing function on the left to the initial selection of \(y_p\) on the right.

\(f(x) = 15 \)
\(y_p = A \)
\(f(x) = x^2 \)
\(y_p = Ax^2 + Bx + C \)
\(f(x) = \cos(2x) \)
\(y_p = A\sin(2x) + B\cos(2x) \)
\(f(x) = 3 e^{6x} \)
\(y_p = Ae^{6x} \)
\(y_p = Ax^2 \)
\(y_p = A\cos(2x) \)

Subsection When Guessing \(y_p\) Works (and When It Doesn’t)

This method of guessing the particular solution based on the forcing function is very effective. However, it does have its limitations. In particular, it only applies when \(f(x)\) can be written as a sum or product of:

Polynomials, like \(f(x) = 3x^2 - 4x + 5\)
Exponentials, like \(f(x) = 6 e^{2x}\)
Sine and cosine functions, like \(f(x) = \cos(3x) + 2\sin(3x)\)
Any sum or product of these function types

Functions such as \(\ln x\text{,}\) \(\tan x\text{,}\) or \(1/x\) fall outside this category. For those, we’ll need a different method entirely. For the allowed types, this method provides a fast and reliable way to select \(y_p\text{.}\)

Checkpoint 154. 📖❓ When to Guess \(y_p\).

Which types of forcing functions \(f(x)\) are suitable for the Method of Undetermined Coefficients?

Polynomial functions like \(3x^2 - 4x + 5\)
Correct! Polynomial functions are suitable for this method.
Exponential functions like \(e^{2x}\)
Correct! Exponential functions are also suitable.
Sine or cosine functions like \(\sin(3x)\) or \(\cos(3x)\)
Correct! Sine and cosine functions are suitable as well.
Functions like \(\tan(x)\) or \(\ln(x)\)
Not quite. The method does not apply to logarithmic or tangent functions.

Subsection Combining Basic Forms via Addition

If \(f(x)\) is a sum of different function types, say a polynomial plus an exponential, then \(y_p\) should be a sum of the corresponding matching forms.

🌌 Example 155. Initial form of \(y_p\).

Use Table 151 to identify the appropriate form of \(y_p\) in each case.

\(\ds y'' + 2y' + 3y = 2e^{x} - 1\)

Solution.

This matches an exponential plus constant, so:

\begin{equation*} y_p = Ae^{x} + B \text{.} \end{equation*}

\(\ds u'' - u = 3\sin(2\theta) + \cos(2\theta)\)

Solution.

Both terms have the same argument, \(2\theta\text{,}\) so the form is the standard row 6.

\begin{equation*} y_p = A\sin(2\theta) + B\cos(2\theta) \text{.} \end{equation*}

\(\ds \omega''' + \omega' = 5e^{3w} - 2\sin(4w)\)

Solution.

This combines exponential and sine, so we need:

\begin{equation*} y_p = Ae^{3w} + B\sin(4w) + C\cos(4w) \text{.} \end{equation*}

\(\ds P' + P = \cos(-3t) - \cos(5t)\)

Solution.

Cosines with different arguments require the sum of two forms:

\begin{align*} y_p = A\sin(3t) \amp + B\cos(3t) \\ \amp + C\sin(5t) + D\cos(5t) \end{align*}

Checkpoint 156. 📖❓ Sum of particular solution forms.

Select the initial form of the particular solution for an LNCC equation with forcing function:

\begin{equation*} f(x) = 3e^{4x} + 2\sin(3x)\text{.} \end{equation*}

\(\ds\quad y_p = A e^{4x} + B e^{3x}\)
Incorrect. Derivatives of this form will not yield a \(\sin(3x)\) term.
\(\ds\quad y_p = A e^{4x} + B \sin(3x)\)
Incorrect. This form is missing a term.
\(\ds\quad y_p = e^{4x}(A \sin(3x) + B \cos(3x))\)
Incorrect. This form is inappropriate because the forcing function is not a product of an exponential function and a sine or cosine function.
\(\ds\quad y_p = A e^{4x} + B \sin(3x) + C \cos(3x)\)
Correct! Although the forcing function does not contain a cosine term, the derivatives of cosine terms can still produce sine terms.

Subsection Combining Basic Forms via Multiplication

If the forcing function \(f(x)\) is a product of basic function types—such as a polynomial times an exponential—then the particular solution \(y_p\) must also be a product of their matching forms. For example, the differential equation

\begin{equation*} y'' + y' + 5y = 10x^2 e^{6x}, \end{equation*}

has a forcing function that is a quadratic times an exponential, so we start with:

\begin{equation*} y_p = (Ax^2 + Bx + C)(D e^{6x})\text{.} \end{equation*}

However, be careful: multiplying by \(D\) just makes new coefficients:

\begin{equation*} y_p = (AD x^2 + BD x + CD) e^{6x} \end{equation*}

Since each pair can be merged as a single coefficient, we just relabel them:

\begin{equation*} y_p = (Ax^2 + Bx + C) e^{6x} \end{equation*}

We’ll use this simplification trick often. It keeps the form of \(y_p\) clean and will make solving for \(A\text{,}\) \(B\text{,}\) and \(C\) much easier later.

Checkpoint 157. 📖❓ Product of particular solution forms.

Select the initial form of the particular solution for an LNCC equation with forcing function:

\begin{equation*} f(x) = x^2 e^{x}\text{.} \end{equation*}

\(\ds\quad y_p = (Ax^2 + Bx + C)e^{x}\)
Correct! The forcing function is the product of a 2nd degree polynomial and an exponential function, so the particular solution should be the product the most general 2nd degree polynomial and an exponential function.
\(\ds\quad y_p = Ax^2e^{x}\)
Incorrect. Although this form perfectly matches the form of the forcing function, the solution could potentially include \(x\) and free terms.
\(\ds\quad y_p = (x^2 + Ax + B)e^x\)
Incorrect. The \(x^2\) term also needs a coefficient.
\(\ds\quad y_p = (Ax^2 + Bx + C)(De^{x})\)
Incorrect. The extra coefficient \(D\) is unnecessary, as it can be absorbed into the constants \(A, B,\) and \(C\text{.}\)

🌌 Example 158. Initial form of \(y_p\) - Products of Forms.

For each of the following forcing functions, select an appropriate initial form of \(y_p\text{.}\) Simplify by combining or eliminating any redundant constants.

\(\ds f(x) = e^{-x}\cos(x)\)

Solution.

This is an exponential times a cosine:

\begin{equation*} y_p = Ae^{-x}(B\sin x + C\cos x)\text{.} \end{equation*}

However, distributing \(A\) shows that it is redundant and we relabel as:

\begin{equation*} y_p = e^{-x}(A\sin x + B\cos x)\text{.} \end{equation*}

\(\ds f(x) = e^{6x}e^{-2x}\)

Solution.

The forcing function simplifies: \(e^{6x}e^{-2x} = e^{4x}\text{,}\) so:

\begin{equation*} y_p = Ae^{4x} \end{equation*}

\(\ds f(t) = t^{3}\cos t\)

Solution.

This is a 3rd-degree polynomial multiplied by a cosine. So, initially we set:

\begin{equation*} y_p = (At^3 + Bt^2 + Ct + D)(E\sin t + F\cos t)\text{.} \end{equation*}

Multiplying the polynomial by sine and cosine gives us

\begin{align*} y_p = (At^3 \amp + Bt^2 + Ct + D)E\sin t \\ \amp + (At^3 + Bt^2 + Ct + D)F\cos t \text{.} \end{align*}

Distributing \(E\) and \(F\) shows that they are redundant, so we relabel as follows:

\begin{align*} y_p = (At^3 \amp + Bt^2 + Ct + D)\sin t \\ \amp + (Et^3 + Ft^2 + Gt + H)\cos t \end{align*}

\(\ds f(x) = x^2\cos x + x \sin x\)

Solution.

Here, both terms involve powers of \(x\) times trig functions. But we don’t need separate guesses—one form covers both:

\begin{align*} y_p = (Ax^2 + Bx \amp + C)\cos x \\ \amp + (Dx^2 + Ex + F)\sin x \end{align*}

This form includes the \(x^2\cos x\) term and also handles \(x\sin x\) automatically.

With these examples and the table as a guide, you’re now equipped to handle both additive and multiplicative combinations in \(f(x)\text{.}\) Next, we’ll deal with one last wrinkle: what to do if \(y_p\) accidentally overlaps with the homogeneous solution \(y_h\text{.}\)

Checkpoint 159. 📖❓ Matching Forcing Function (with Products) to \(y_p\).

By dragging from left to right, match the forcing functions to the appropriate initial form of \(y_p\text{.}\)

\(f(x) = x e^{-12x} \)
\(y_p = (Ax + B)e^{-12x} \)
\(y_p = Ax e^{-12x} \)
\(y_p = A e^{-12x} \cos x \)
\(f(x) = x \sin x \)
\(y_p = (Ax + B)\cos x + (Cx + D)\sin x\)
\(f(x) = e^{-12x} \cos x \)
\(y_p = e^{-12x}(A\cos x + B\sin x) \)
\(y_p = A x \sin x \)
\(y_p = (Ax + B) \sin x \)

Subsection Modifying Particular Solutions

We now know how to choose a form for \(y_p\) based on the structure of \(f(x)\text{,}\) but there’s one final complication: If a term in \(y_p\) matches a term in the homogeneous solution \(y_h\text{,}\) then that term cannot contribute to generating \(f(x)\text{.}\) Instead, it will give zero, since that is what terms in \(y_h\) do when substituted into the differential equation.

To fix this, we multiply the overlapping part of \(y_p\) by \(x\text{.}\) If there’s still overlap, we multiply by \(x\) again—repeating until \(y_p\) shares no like terms with \(y_h\text{.}\)

For example, consider:

\begin{equation*} y'' - 4y' + 3y = e^{3x} \end{equation*}

The homogeneous solution and the initial particular solution are:

\begin{equation*} y_h = c_1 e^x + {\DLO c_2 e^{3x}}, \quad y_p = {\DLO A e^{3x}}\text{.} \end{equation*}

Notice that \(y_h\) and \(y_p\) both contain the like terms: \(c_2 e^{3x}\) and \(A e^{3x}\text{.}\) This means that plugging \(y_p\) into the left side of the equation will simplify to \(0\text{,}\) not \(e^{3x}\text{,}\) as required.

So we modify \(y_p\) by multiplying the conflicting term by \(x\text{:}\)

\begin{equation*} y_p = A x e^{3x} \end{equation*}

Now, \(y_p\) has no terms in common with \(y_h\text{,}\) ensuring that they are independent.

Modification Rule for the Particular Solution.

If any term in your guess for \(y_p\) also appears in \(y_h\text{,}\) multiply that term by the independent variable \(x\text{.}\) Repeat as needed until all overlapping terms are eliminated.

Checkpoint 160. 📖❓ Handling Like Terms.

What should you do if \(y_p\) contains a term also present in \(y_h\text{?}\)

Multiply the term in \(y_p\) by the independent variable.
Delete the term from \(y_p\text{.}\)
Multiply the term in \(y_h\) by the independent variable.
Use integration to eliminate the term.

Let’s work through some examples where modification is required, and see how this rule plays out.

🌌 Example 161. Modifying the Initial Form of \(y_p\).

For each LNCC equation below:

Find the homogeneous solution \(y_h\text{.}\)
Select the initial form of \(y_p\text{.}\)
Modify \(y_p\) to ensure it has no overlap with \(y_h\text{.}\)

\(\ds y'' - 2y' + y = e^x\)

Solution.

The characteristic equation is

\begin{equation*} r^2 - 2r + 1 = 0\text{,} \end{equation*}

which has a repeated root: \(r=1\text{.}\) So,

\begin{equation*} y_h = c_1 e^x + c_2 x e^x \text{.} \end{equation*}
The natural guess for \(y_p\) to match \(e^x\) is

\begin{equation*} y_p = A e^x\text{.} \end{equation*}
Since \(e^x\) is in \(y_h\text{,}\) we multiply \(y_p\) by \(x\text{:}\)

\begin{equation*} y_p = A x e^x \end{equation*}

But, \(x e^x\) is still in \(y_h\text{,}\) so we repeat:

\begin{equation*} y_p = A x^2 e^x\text{.} \end{equation*}

\(y''' + 4y' = \cos(2x) - \sin(x)\)

Solution.

Since the characteristic equation is

\begin{equation*} r^3 + 4r = 0\text{,} \end{equation*}

the roots are: \(r = 0, \pm 2i\) and

\begin{equation*} y_h = c_1 + c_2 \cos(2x) + c_3 \sin(2x)\text{.} \end{equation*}
The guess for \(y_p\) is

\begin{align*} y_p \amp = \ob{A \cos(2x) + B \sin(2x)}^{\large\text{already in } y_h}\\ \amp \hphantom{=\ A \cos(2x)} + D \cos(x) + E \sin(x)\text{.} \end{align*}
Only the \(\cos(2x)\) and \(\sin(2x)\) terms conflict. Modify them:

\begin{align*} y_p = Ax\cos(2x) \amp + Bx\sin(2x) \\ \amp + D \cos(x) + E \sin(x)\text{.} \end{align*}

\(y'' + y = x e^x\)

Solution.

The characteristic equation

\begin{equation*} r^2 + 1 = 0\text{,} \end{equation*}

has roots: \(r = \pm i\text{,}\) leading to

\begin{equation*} y_h = c_1 \cos(x) + c_2 \sin(x)\text{.} \end{equation*}
The initial guess for \(y_p\) is

\begin{equation*} y_p = (Ax + B)\ e^x\text{.} \end{equation*}
No modification is needed since \(e^x\) doesn’t appear in \(y_h\text{.}\)

\(y'' - 6y' + 9y = x^2 e^{3x}\)

Solution.

The characteristic equation

\begin{equation*} r^2 - 6r + 9 = 0\text{,} \end{equation*}

has the repeated solution \(r = 3\text{,}\) so

\begin{equation*} y_h = (c_1 + c_2 x) e^{3x}\text{.} \end{equation*}
The initial guess for \(y_p\) is:

\begin{equation*} y_p = (Ax^2 + Bx + C) e^{3x}\text{.} \end{equation*}
Multiplying out \(y_h\) and \(y_p\) helps show that \(e^{3x}\) and \(x e^{3x}\) are conflicting terms. Multiplying all terms by \(x^2\) gives the correct form:

\begin{equation*} y_p = (Ax^4 + Bx^3 + Cx^2) e^{3x}\text{.} \end{equation*}

Now that we know how to construct an appropriate and independent form for \(y_p\text{,}\) the last step is to determine the values of the unknown coefficients.

Checkpoint 162. 📖❓ Select the Correct Form of \(y_p\).

Consider the linear non-homogeneous constant coefficient equation

\begin{equation*} y'' - 4y' + 3y = \frac12 e^x\text{.} \end{equation*}

Given that the homogeneous solution is \(y_h = c_1 e^x + c_2 e^{3x}\text{,}\) select the form of the particular solution after any necessary modifications.

\(\quad y_p = Ae^x \)
Incorrect. This would be the correct initial form for the particular solution, but not the final form.
\(\quad y_p = Ae^x + Be^{3x} \)
Incorrect. Don’t create the particular solution form based on the homogeneous solution.
\(\quad y_p = (Ax + B)e^x\)
Incorrect. This form indicates the forcing function contains the product of a first degree polynomial and an exponential function.
\(\quad y_p = Axe^{x} \)
Correct! The initial form would be \(Ae^x\text{,}\) but a like-term is also in \(y_h\text{.}\) One multiplication by \(x\) resolves this.

Checkpoint 163. 🤔💭 Reading Questions.

(a) 🤔💭 Match Forcing Function to \(y_p\) Form.

Match each forcing function \(f(x)\) to its correct initial form of \(y_p\text{.}\)

\(f(x) = x^3\)
\(y_p = A x^3 + B x^2 + C x + D\)
\(f(x) = e^{2x}\)
\(y_p = A e^{2x}\)
\(f(x) = 2x + 5\)
\(y_p = A x + B\)
\(f(x) = e^{3x}\)
\(y_p = A e^{3x}\)
\(f(x) = \cos(4x)\)
\(y_p = A \cos(4x) + B \sin(4x)\)
\(f(x) = x^2 e^x\)
\(y_p = (A x^2 + B x + C) e^x\)
\(f(x) = x \cos x\)
\(y_p = (A x + B) \cos x + (C x + D) \sin x\)

(b) 🤔💭 What is the Next Move?

Which statement best describes the next action you should take when the particular solution, \(y_p\text{,}\) has a like-term with the homogeneous solution, \(y_h\text{?}\)

This is normal, move on.
Incorrect. The \(y_p\) cannot have terms in common with \(y_h\text{.}\)
Remove the like-term from \(y_p\text{.}\)
Incorrect. No terms in \(y_h\) should ever be removed.
Multiply the like-term in \(y_p\) by \(x\text{.}\)
Correct!
Multiply the like-term in \(y_h\) by \(x\text{.}\)
Incorrect. The particular solution should be modified, not the homogeneous solution.

(c) Modify \(y_p\).

If an equation has a forcing function and a homogeneous solution,

\begin{equation*} f(x) = x^2 e^{3x}\quad \text{and} \quad y_h = (c_1 + c_2 x) e^{3x}\text{,} \end{equation*}

select the correct modified form of \(y_p\text{?}\)

\(\quad y_p = A x^2 e^{3x}\)
Incorrect. The form of \(y_p\) should be modified until there are no terms in common with \(y_h\text{.}\)
\(\quad y_p = (A x^4 + B x^3 + C x^2) e^{3x}\)
Correct! The modified form of \(y_p\) is \((A x^4 + B x^3 + C x^2) e^{3x}\text{.}\)
\(\quad y_p = (A x^3 + B x^2) e^{3x}\)
Incorrect. The form of \(y_p\) should be modified until there are no terms in common with \(y_h\text{.}\)
\(\quad y_p = (A x^2 + B x) e^{3x}\)
Incorrect. The form of \(y_p\) should be modified until there are no terms in common with \(y_h\text{.}\)

(d) 🤔💭 Match the Scenario to the Correct Action.

The homogeneous solution to the LNCC equation

\begin{equation*} y'' - 2y' + y = f(x) \end{equation*}

is given by

\begin{equation*} y_h = c_1 e^{2x} + c_2 x e^{2x}\text{.} \end{equation*}

On the left side are initial \(y_p\) forms for various forcing functions, \(f(x)\text{.}\)

Drag each initial \(y_p\) form to its correct modified form on the right.

\(y_p = e^{2x}\left(A\cos x + B\sin x\right) \)
\(y_p = Ae^{4x} \)
No modification needed
\(y_p = (Ax^3+Bx^2+Cx)e^{2x}\)
\(y_p = (Ax+B)e^{2x} \)
\(y_p = (Ax^3+Bx^2)e^{2x}\)
\(y_p = Axe^{2x} \)
\(y_p = (Ax^2+Bx+C)e^{2x}\)
\(y_p = Ae^{2x} \)
\(y_p = Ax^2e^{2x} \)
\(y_p = (Ax^2+Bx)e^{2x}\)

(e) 🤔💭 REVIEW: Valid Forcing Functions for Undetermined Coefficients.

Which of the following are valid types of forcing functions for the Method of Undetermined Coefficients?

\(f(x) = 5x^2 + 1\)
\(f(x) = 3e^{2x} + \cos(4x)\)
\(f(x) = \ln x\)
\(f(x) = \tan(x)\)
\(f(x) = x \sin(3x)\)

You have attempted of activities on this page.

Prev Top Next