DE-BOOK Selecting Particular Solutions

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:

a_{n} y^{(n)} + \dots + a_{2} y^{″} + a_{1} y^{'} + a_{0} y = f (x) .

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} .

The form of this solution depends on the type of function

f (x)

represents. To help select the correct form of

y_{p},

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) .

$f (x)$	Particular Solution Form, $y_{p}$
$a$	$A$
$a x + b$	$A x + B$
$a x^{2} + b x + c$	$A x^{2} + B x + C$
$a x^{3} + b x^{2} + c x + d$	$A x^{3} + B x^{2} + C x + D$
$a e^{α x}$	$A e^{α x}$
$a \sin (β x)$	$A \sin (β x) + B \cos (β x)$
$a \cos (β x)$	$A \sin (β x) + B \cos (β x)$

In this table, the constants

a,

b,

c,

d,

α,

and

β

are assumed to be known, while

A,

B,

C,

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. Determine the initial form of $y_{p}$ .

Solution 1.

y^{″} + y^{'} + 5 y = 3 e^{6 x}

Since

f (x) = 3 e^{6 x}

has the form

a e^{α x},

we set

y_{p} = A e^{6 x} .

Solution 2.

y^{″} + 3 y^{'} - y = 2 \sin (2 x)

2 \sin (2 x)

has the form

a \sin (β x),

y_{p} = A \sin (2 x) + B \cos (2 x) .

Solution 3.

ω^{‴} + ω^{'} = 5 θ^{2} + θ + 1

5 θ^{2} + θ + 1

has the form

a θ^{2} + b θ + c,

ω_{p} = A θ^{2} + B θ + C .

Solution 4.

2 \frac{d y}{d x} + 5 y = - 3

Since

f (x) = - 3

is of the form

a,

we set

y_{p} = A .

Solution 5.

y^{″} - 3 y^{'} + 0.5 y = 14 x^{4}

Even though

f (x) = 14 x^{4}

only has an

x^{4}

term, we consider it to be of the form

a x^{4} + b x^{3} + c x^{2} + d x + e,

where

b = c = d = e = 0 .

Therefore,

y_{p} = A x^{4} + B x^{3} + C x^{2} + D x + E .

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) .

Let’s look at an example to clarify this approach.

Example 17. Determine the initial form of $y_{p}$ .

For some LNCC equation of the form:

a y^{‴} + b y^{″} + c y^{'} + d y = f (w),

give the form of

y_{p}

for each

f (w)

below.

Solution 1.

f (w) = 2 e^{w} - 1

Since

f (w)

is the sum of an

e^{w}

term and a constant term, we set

y_{p} = A e^{w} + B .

Solution 2.

f (w) = 3 \sin (2 w) + 4 \cos (2 w)

Since

f (w)

is the sum of a

\sin (2 w)

term and a

\cos (2 w)

term, we set

y_{p} = A \sin (2 w) + B \cos (2 w) .

Solution 3.

f (w) = 5 e^{3 w} - 2 \sin (4 w)

Since

f (w)

is the sum of an

e^{3 w}

term and a

\sin (4 w)

term, we set

y_{p} = A e^{3 w} + B \sin (4 w) + C \cos (4 w) .

Solution 4.

f (w) = 2 \sin (- 3 w) - 4 \cos (5 w)

Since

f (w)

is the sum of a

\sin (- 3 w)

term and a

\cos (5 w)

term, we set

y_{p} = A \sin (- 3 w) + B \cos (- 3 w) + C \sin (5 w) + D \cos (5 w) .

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. Find the initial form of $y_{p}$ for the given equation.

Solution.

y^{″} + y^{'} + 5 y = 10 x^{2} e^{6 x}

Since

f (x) = 10 x^{2} e^{6 x}

is the product of a polynomial and an exponential, we initially set:

y_{p} = (A x^{2} + B x + C) (D e^{6 x}) .

However,

D

is redundant because if we multiply it onto the polynomial, it can be absorbed into

A, B,

and

C,

as follows:

y_{p} = (A x^{2} + B x + C) (D e^{6 x}) = (A D x^{2} + B D x + C D) e^{6 x} .

So, the form of

y_{p}

simplifies to:

y_{p} = (A x^{2} + B x + C) e^{6 x} .

Here are a few tips to help you avoid redundant coefficients when constructing the form of

y_{p} .

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:
$(polynomial form) \sin (β x) + (polynomial form) \cos (β x)$
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. Find the initial form of $y_{p}$ for each equation.

Solution 1.

y^{″} + y = e^{- x} \cos (x)

Since

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

is the product of an exponential and a cosine function, we apply tip 1 and set:

y_{p} = e^{- x} (A \sin x + B \cos x) .

Solution 2.

y^{″} - 4 y = e^{6 x} e^{- 3 x}

Since

f (x) = e^{6 x} e^{- 3 x}

is the product of two exponential functions, we ignore one of the coefficients and set:

y_{p} = A e^{6 x} e^{- 3 x} .

Note,

e^{6 x} e^{- 3 x}

is the same as

e^{3 x},

so we could also write

y_{p} = A e^{3 x} .

Solution 3.

9 P^{'} - P = t^{3} \cos (t)

Since

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

is the product of a polynomial and a cosine function, we apply tip 2 and set:

P_{p} = (A t^{3} + B t^{2} + C t + D) \sin t + (E t^{3} + F t^{2} + G t + H) \cos t .

Solution 4.

2 y^{″} + 3 y^{'} - 15 y = x^{2} \cos x + x \sin x

When setting up the particular solution, it’s easy to introduce redundant terms if we aren’t careful. For example, if we initially set:

\begin{matrix} y_{p} = (A x^{2} + \underset{―}{B x + C) \sin x} + (D x^{2} + \underset{―}{E x + F) \cos x} \end{matrix}

and

\begin{matrix} (\underset{―}{G x + H) \sin x} + (\underset{―}{I x + J) \cos x}, \end{matrix}

you may notice that the underlined terms introduce unnecessary repetition. By grouping like terms, we can simplify this to:

y_{p} = (A x^{2} + B x + C) \sin x + (D x^{2} + E x + F) \cos x .

Reading Questions Check-Point Questions

1. Which of the following functions would be an appropriate form for the particular solution $y_{p}$ if $f (x) = 5 x^{2} + 3 x + 1 ?$

y_{p}

f (x) = 5 x^{2} + 3 x + 1 ?

$y_{p} = A x^{2} + B x + C$
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.
$y_{p} = A x^{3} + B x^{2} + C x + D$
Incorrect. This form would be too complex, as the given function is quadratic, not cubic.

2. If $f (x) = 3 e^{4 x} + 2 \sin (3 x),$ what should be the form of the particular solution $y_{p} ?$

f (x) = 3 e^{4 x} + 2 \sin (3 x),

y_{p} ?

$y_{p} = A e^{4 x} + B \sin (3 x) + C \cos (3 x)$
Correct! The particular solution should account for both the exponential and trigonometric terms in $f (x) .$
$y_{p} = A e^{4 x} + B e^{3 x}$
Incorrect. The second term should include sine and cosine, not another exponential.
$y_{p} = A \sin (3 x) + B \cos (3 x)$
Incorrect. This form only accounts for the trigonometric part of $f (x),$ not the exponential.
$y_{p} = A e^{4 x} \sin (3 x)$
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 $f (x) = x^{2} e^{x} ?$

$y_{p} = (A x^{2} + B x + C) e^{x}$
Correct! The particular solution should combine the polynomial with the exponential function in the same way as $f (x) .$
$y_{p} = A e^{x}$
Incorrect. This form would be appropriate for an exponential function, but it doesn’t account for the polynomial $x^{2} .$
$y_{p} = A x^{2} + B x + C$
Incorrect. This form only accounts for the polynomial part of $f (x),$ not the exponential part.
$y_{p} = (A x^{2} + B x + C) (D e^{x})$
Incorrect. The extra coefficient $D$ is unnecessary, as it can be absorbed into the constants $A, B,$ and $C .$

4. Match the $y_{p}$ form.

Suppose we want the form of particular solution,

y_{p},

for the equation

y^{″} + y^{'} + 5 y = f (x) .

On the left are possible

f (x)

functions. Drag each of these functions to the correct form of

y_{p}

that you would use.

$15$
$A$
$x^{2}$
$A x^{2} + B x + C$
$\cos (2 x)$
$A \sin (2 x) + B \cos (2 x)$
$\sin x$
$A \sin (x) + B \cos (x)$
$3 e^{6 x}$
$A e^{6 x}$

5. Match the $y_{p}$ form.

Suppose we want the form of particular solution,

y_{p},

for the equation

y^{″} + y^{'} + 5 y = f (x) .

Drag the form of

y_{p}

that you would use for each function,

f (x) .

$x e^{2 x}$
$(A x + B) e^{2 x}$
$x^{2} \cos (3 x)$
$(A x^{2} + B x + C) \cos (3 x)$
$+ (D x^{2} + E x + F) \sin (3 x)$
$e^{x} \sin (4 x)$
$e^{x} (A \sin (4 x) + B \cos (4 x))$
$x^{2} e^{- x}$
$(A x^{2} + B x + C) e^{- x}$
$x \cos (4 x)$
$(A x + B) \cos (4 x)$
$+ (C x + D) \sin (4 x)$
$e^{x} \sin (x)$
$e^{x} (A \sin (x) + B \cos (x))$
$x^{3} e^{2 x}$
$(A x^{3} + B x^{2} + C x + D) e^{2 x}$
$x e^{x} \cos (x)$
$e^{x} (A x + B) \cos (x)$
$+ e^{x} (C x + D) \sin (x)$
$x^{2} \sin (2 x)$
$(A x^{2} + B x + C) \sin (2 x)$
$+ (D x^{2} + E x + F) \cos (2 x)$
$e^{3 x} \cos (2 x)$
$e^{3 x} (A \cos (2 x) + B \sin (2 x))$

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Section 7.3 Selecting Particular Solutions

Example 16. Determine the initial form of yp.

Example 17. Determine the initial form of yp.

Example 18. Find the initial form of yp for the given equation.

Tips to Avoid Redundant Coefficients.

Example 19. Find the initial form of yp for each equation.

Reading Questions Check-Point Questions

1. Which of the following functions would be an appropriate form for the particular solution yp if ?f(x)=5x2+3x+1?

2. If ,f(x)=3e4x+2sin⁡(3x), what should be the form of the particular solution ?yp?

3. Which of the following would be the correct form for yp if ?f(x)=x2ex?

4. Match the yp form.

5. Match the yp form.

Example 16. Determine the initial form of $y_{p}$ .

Example 17. Determine the initial form of $y_{p}$ .

Example 18. Find the initial form of $y_{p}$ for the given equation.

Example 19. Find the initial form of $y_{p}$ for each equation.

1. Which of the following functions would be an appropriate form for the particular solution $y_{p}$ if $f (x) = 5 x^{2} + 3 x + 1 ?$

2. If $f (x) = 3 e^{4 x} + 2 \sin (3 x),$ what should be the form of the particular solution $y_{p} ?$

3. Which of the following would be the correct form for $y_{p}$ if $f (x) = x^{2} e^{x} ?$

4. Match the $y_{p}$ form.

5. Match the $y_{p}$ form.