Section 7.3 Selecting Particular Solutions
As discussed in the previous section, solving non-homogeneous linear differential equations involves finding a solution that satisfies the equation:
\begin{equation*}
a_n y^{(n)} + \cdots + a_2 y'' + a_1 y' + a_0 y = f(x)\text{.}
\end{equation*}
The challenge is that the terms on the left side need to combine in just the right way to match \(f(x)\) on the right. This means the form of the particular solution must resemble \(f(x)\) closely. When \(f(x)\) contains polynomial, exponential, or trigonometric (sine or cosine) terms, the particular solution should involve similar components.
The specific function that accounts for
\(f(x)\) on the right side is known as the
particular solution , denoted
\(y_p\text{.}\) The form of this solution depends on the type of function
\(f(x)\) represents. To help select the correct form of
\(y_p\text{,}\) refer to
Table 15 , which outlines the most common types of
\(f(x)\) and their corresponding
\(y_p\) forms.
Table 15. Selecting the form of \(y_p\) based on the form of \(f(x)\text{.}\)
\(f(x)\)
Particular Solution Form, \(y_p\)
\(a\) \(A\)
\(ax+b\) \(Ax+B\)
\(ax^2+bx+c\) \(Ax^2+Bx+C\)
\(ax^3+bx^2+cx+d\) \(Ax^3+Bx^2+Cx+D\)
\(a e^{\alpha x}\) \(A e^{\alpha x}\)
\(a \sin(\beta x)\) \(A \sin(\beta x) + B \cos(\beta x)\)
\(a \cos(\beta x)\) \(A \sin(\beta x) + B \cos(\beta x)\)
In this table, the constants \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) \(d\text{,}\) \(\alpha\text{,}\) and \(\beta\) are assumed to be known, while \(A\text{,}\) \(B\text{,}\) \(C\text{,}\) etc., need to be determined. We will explore how to determine these coefficients in the next section. For now, we are concerned with picking the correct form. Let’s work through a few examples to solidify this process.
Example 16 .
\(\ \ \) \(y_p\)
\(\ds y'' + y' + 5y = 3e^{6x} \)
Solution .
\(f(x) = 3e^{6x}\) \(a e^{\alpha x}\text{,}\)
\begin{equation*}
y_p = Ae^{6x}\text{.}
\end{equation*}
\(\ds y'' + 3y' - y = 2\sin(2x) \)
Solution .
\(2\sin(2x)\) \(a \sin(\beta x)\text{,}\)
\begin{equation*}
y_p = A\sin(2x) + B\cos(2x)\text{.}
\end{equation*}
\(\ds \omega''' + \omega' = 5\theta^2 + \theta + 1 \)
Solution .
\(5\theta^2 + \theta + 1\) \(a\theta^2+b\theta+c\text{,}\)
\begin{equation*}
\omega_p = A\theta^2 + B\theta + C\text{.}
\end{equation*}
\(\ds 2\frac{dy}{dx} + 5y = -3 \)
Solution .
\(f(x) = -3\) \(a\text{,}\)
\begin{equation*}
y_p = A\text{.}
\end{equation*}
\(\ds y'' -3y' + 0.5y = 14x^4 \)
Solution .
\(f(x) = 14x^4\) \(x^4\)
\begin{equation*}
ax^4+bx^3+cx^2+dx+e\text{,}
\end{equation*}
\(b = c = d = e = 0\text{.}\)
\begin{equation*}
y_p = Ax^4+Bx^3+Cx^2+Dx+E\text{.}
\end{equation*}
When \(f(x)\) is a sum of multiple functions, each part of the sum contributes to the form of the particular solution. In these situations, we construct \(y_p\) by simply adding together the particular solutions corresponding to each term in \(f(x)\text{.}\) Let’s look at an example to clarify this approach.
Example 17 .
\(\ \ \) \(f\text{,}\) \(y_p\)
Solution .
\(f(w)\) \(e^{w}\)
\begin{equation*}
y_p = Ae^{w} + B\text{.}
\end{equation*}
\(\ds 3e^{6x} + 2\sin(2x) \)
Solution .
\(f(x)\) \(e^{6x}\) \(\sin(2x)\)
\begin{equation*}
y_p = Ae^{6x} + B\sin(2x) + C\cos(2x)\text{.}
\end{equation*}
\(\ds 2\sin(-2x) - 6\cos(3x) \)
Solution .
\(f(x)\) \(\sin(-2x)\) \(\cos(3x)\)
\begin{align*}
y_p = A\sin(-2x) \amp\ + B\cos(-2x) \\
+ \amp\ C\sin(3x) + D\cos(3x) \text{.}
\end{align*}
When \(f(x)\) involves products of functions, constructing \(y_p\) can become more involved. This is because multiplying terms may introduce extra coefficients that turn out to be unnecessary or redundant. In these cases, simplifying the solution to avoid duplication is essential. The following example illustrates this process.
Example 18 .
\(\ \ \) \(y_p\)
\begin{equation*}
y'' + y' + 5y = 10x^2 e^{6x}\text{.}
\end{equation*}
Solution .
Since \(f(x) = 10x^2 e^{6x}\) is the product of a polynomial and an exponential, we initially set:
\begin{equation*}
y_p = (Ax^2 + Bx + C)(D e^{6x})\text{.}
\end{equation*}
However, \(D\) is redundant because if we multiply it onto the polynomial, it can be absorbed into \(A, B, \) and \(C\text{,}\) as follows:
\begin{equation*}
y_p = (Ax^2 + Bx + C)(D e^{6x}) = \left(AD x^2 + BD x + CD\right) e^{6x}\text{.}
\end{equation*}
So, the form of \(y_p\) simplifies to:
\begin{equation*}
y_p = \left(A x^2 + B x + C\right) e^{6x}\text{.}
\end{equation*}
Here are a few tips to help you avoid redundant coefficients when constructing the form of \(y_p\text{.}\)
Tips to Avoid Redundant Coefficients.
When \(f(x)\) is the product of an exponential and another function, avoid writing a coefficient for the exponential term.
For products of polynomials and sine/cosine functions, use:
\begin{equation*}
(\text{polynomial form}) \sin(\beta x) + (\text{polynomial form}) \cos(\beta x)
\end{equation*}
where the polynomial form on each term uses different coefficients.
When working with sums, look for redundant overlapping terms.
Let’s put these tips into practice with a couple of examples.
Example 19 .
\(\ \ \) \(y_p\)
\(\ds y'' + y = e^{-x}\cos(x) \) Solution .
\(f(x) = e^{-x}\cos(x)\)
\begin{equation*}
y_p = e^{-x}(A\sin x + B\cos x)\text{.}
\end{equation*}
\(\ds y'' - 4y = e^{6x}e^{-3x} \) Solution .
\(f(x) = e^{6x}e^{-3x}\)
\begin{equation*}
y_p = Ae^{6x}e^{-3x}\text{.}
\end{equation*}
\(e^{6x}e^{-3x}\) \(e^{3x}\text{,}\) \(y_p = Ae^{3x}\text{.}\)
\(\ds 9P' - P = t^{3}\cos(t) \) Solution .
\(f(t) = t^{3}\cos(t)\)
\begin{equation*}
P_p = (At^3 + Bt^2 + Ct + D)\sin t + (Et^3 + Ft^2 + Gt + H)\cos t\text{.}
\end{equation*}
\(\ds 2y'' + 3y' - 15y = x^2\cos x + x \sin x \) Solution .
\begin{gather*}
y_p = (Ax^2 + \ul{Bx + C) \sin x} + (Dx^2 + \ul{Ex + F)\cos x}
\end{gather*}
\begin{gather*}
(\ul{Gx + H)\sin x} + (\ul{Ix + J)\cos x},
\end{gather*}
\begin{equation*}
y_p = (Ax^2 + Bx + C)\sin x + (Dx^2 + Ex + F)\cos x\text{.}
\end{equation*}
Reading Questions Check-Point Questions
1. Which of the following functions would be an appropriate form for the particular solution \(y_p\) if \(f(x) = 5x^2 + 3x + 1\text{?}\)
\(y_p = Ax^2 + Bx + C\) Correct! The form of the particular solution for a quadratic polynomial should match the degree of the polynomial.
\(y_p = A e^{x} + B\) Incorrect. This form would be appropriate if \(f(x)\) was an exponential function, not a polynomial.
\(y_p = A \sin(x) + B \cos(x)\) Incorrect. This form would be appropriate if \(f(x)\) was a trigonometric function.
\(y_p = Ax^3 + Bx^2 + Cx + D\) Incorrect. This form would be too complex, as the given function is quadratic, not cubic.
2. If \(f(x) = 3e^{4x} + 2\sin(3x)\text{,}\) what should be the form of the particular solution \(y_p\text{?}\)
\(y_p = A e^{4x} + B \sin(3x) + C \cos(3x)\) Correct! The particular solution should account for both the exponential and trigonometric terms in \(f(x)\text{.}\)
\(y_p = A e^{4x} + B e^{3x}\) Incorrect. The second term should include sine and cosine, not another exponential.
\(y_p = A \sin(3x) + B \cos(3x)\) Incorrect. This form only accounts for the trigonometric part of \(f(x)\text{,}\) not the exponential.
\(y_p = A e^{4x}\sin(3x)\) Incorrect. This form is not appropriate, as the terms in \(f(x)\) are separate, not a product of exponential and trigonometric terms.
3. Which of the following would be the correct form for \(y_p\) if \(f(x) = x^2 e^{x}\text{?}\)
\(y_p = (Ax^2 + Bx + C)e^{x}\) Correct! The particular solution should combine the polynomial with the exponential function in the same way as \(f(x)\text{.}\)
\(y_p = Ae^{x}\) Incorrect. This form would be appropriate for an exponential function, but it doesn’t account for the polynomial \(x^2\text{.}\)
\(y_p = Ax^2 + Bx + C\) Incorrect. This form only accounts for the polynomial part of \(f(x)\text{,}\) not the exponential part.
\(y_p = (Ax^2 + Bx + C)(De^{x})\) Incorrect. The extra coefficient \(D\) is unnecessary, as it can be absorbed into the constants \(A, B,\) and \(C\text{.}\)
4. Match the \(y_p\) form .
Suppose we want the form of particular solution, \(y_p\text{,}\) for the equation
\begin{equation*}
y'' + y' + 5y = f(x)\text{.}
\end{equation*}
On the left are possible \(f(x)\) functions. Drag each of these functions to the correct form of \(y_p\) that you would use.
\(15 \) \(A \) \(x^2 \) \(Ax^2 + Bx + C \) \(\cos(2x) \) \(A\sin(2x) + B\cos(2x) \) \(\sin x \) \(A\sin(x) + B\cos(x) \) \(3 e^{6x} \) \(Ae^{6x} \)
5. Match the \(y_p\) form .
Suppose we want the form of particular solution, \(y_p\text{,}\) for the equation
\begin{equation*}
y'' + y' + 5y = f(x)\text{.}
\end{equation*}
Drag the form of \(y_p\) that you would use for each function, \(f(x)\text{.}\)
\(x e^{2x} \) \(\scriptsize (Ax + B)e^{2x} \) \(x^2 \cos(3x) \)
\(\scriptsize (Ax^2 + Bx + C)\cos(3x) \) \(\scriptsize \hspace{3.2em} + (Dx^2 + Ex + F)\sin(3x) \)
\(e^{x} \sin(4x) \) \(\scriptsize e^{x}(A\sin(4x) + B\cos(4x)) \) \(x^2 e^{-x} \) \(\scriptsize (Ax^2 + Bx + C)e^{-x} \) \(x \cos(4x) \)
\(\scriptsize (Ax + B)\cos(4x) \) \(\scriptsize \hspace{5.2em} + (Cx + D)\sin(4x) \)
\(e^{x} \sin(x) \) \(\scriptsize e^{x}(A\sin(x) + B\cos(x)) \) \(x^3 e^{2x} \) \(\scriptsize (Ax^3 + Bx^2 + Cx + D)e^{2x} \) \(x e^{x} \cos(x) \)
\(\scriptsize e^{x}(Ax + B)\cos(x) \) \(\scriptsize \hspace{5.5em} + e^{x}(Cx + D)\sin(x) \)
\(x^2 \sin(2x) \)
\(\scriptsize (Ax^2 + Bx + C)\sin(2x) \) \(\scriptsize \hspace{3.2em} + (Dx^2 + Ex + F)\cos(2x) \)
\(e^{3x} \cos(2x) \) \(\scriptsize e^{3x}(A\cos(2x) + B\sin(2x)) \) You have attempted
of
activities on this page.
