DE-BOOK Method of Undetermined Coefficients

Section 7.5 Method of Undetermined Coefficients

When solving a non-homogeneous linear differential equation, selecting the correct form for the particular solution,

y_{p},

is just the beginning. The next step is determining the values of its unknown coefficients, such as

A

and

B .

In this section, we’ll explore how to compute these coefficients by substituting the guessed particular solution into the original equation.

Let’s revisit the example we discussed earlier. Consider the differential equation:

\begin{matrix} (38) & y^{″} - 4 y^{'} + 3 y = 9 x . \end{matrix}

We guessed that the particular solution would be linear, so we set

y_{p} = A x + B .

Our goal now is to determine the values of

A

and

B .

To do this, we substitute

y_{p} = A x + B

into the original differential equation. Here’s how it works:

\begin{matrix} y_{p} = A x + B ⟹ y_{p}^{'} = A ⟹ y_{p}^{″} = 0 \end{matrix}

Substituting

y_{p},

y_{p}^{'},

and

y_{p}^{″}

into (38) results in:

\begin{aligned} 0 - 4 A + 3 (A x + B) = & 9 x \\ - 4 A + 3 A x + 3 B = & 9 x \end{aligned}

By matching the coefficients of like terms on opposite sides of the equation,

we solve:

\begin{aligned} 3 A = & 9 ⟹ A = 3 \\ - 4 A + 3 B = & 0 ⟹ B = 4 \end{aligned}

Thus, the particular solution is:

y_{p} = 3 x + 4 .

Regardless of

y_{p}

’s form, the process of finding the coefficients remains consistent:

Substitute the chosen $y_{p}$ into the differential equation.
Collect and match coefficients of like terms on both sides of the equation.
Solve the resulting system to find $A, B, C, \dots .$

This is why the method is called the "method of undetermined coefficients." and the complete process is summarized as follows.

Method 5. Undetermined Coefficients.

The linear nonhomogeneous equation with constant coefficients,

\begin{matrix} (39) & a_{n} y^{(n)} + a_{n - 1} y^{(n - 1)} + \dots + a_{2} y^{″} + a_{1} y^{'} + a_{0} y = f (x), \end{matrix}

has a general solution given by

\begin{matrix} (40) & y = y_{h} + y_{p}, \end{matrix}

where

y_{h}

is the solution to the homogeneous equation

\begin{matrix} (41) & a_{n} y^{(n)} + a_{n - 1} y^{(n - 1)} + \dots + a_{2} y^{″} + a_{1} y^{'} + a_{0} y = 0, \end{matrix}

and

y_{p}

is the particular solution found through the following steps...

Select Initial $y_{p}$: Select the initial form of $y_{p}$ that generalizes $f (x) .$
Adjust $y_{p}$: If $y_{p}$ shares terms with $y_{h},$ repeatedly multiply the terms by $x$ until there are no terms in common.
Find the Coefficients of $y_{p}$: Plug $y_{p}$ into (39) and match terms on each side to solve for the unknown coefficients of $y_{p} .$

Now for a few examples to illustrate the complete application of the nndetermined coefficients method.

Example 22. Find the general solution for each equation.

Solution 1.

y^{″} - 3 y^{'} + y = 2 x^{2} + 3 x

Following the method of Undetermined Coefficients, we first find

y_{h}

y_{h} = c_{1} e^{x} + c_{2} x e^{x} .

Next, we select the initial form of

y_{p}

\begin{matrix} y_{p} = A x^{2} + B x + C \end{matrix}

and adjust

y_{p}

, if necessary. A quick check shows that

y_{p}

has no terms in common with

y_{h} .

Finally, we find the coefficients of

y_{p}

. To do this, let’s compute

y_{p}^{'},

y_{p}^{″},

plug these into the equation, and collect the

x^{2},

x,

and constant terms as follows:

y_{p}^{'} = 2 A x + B y_{p}^{″} = 2 A

\begin{matrix} 2 A - 3 (2 A x + B) + (A x^{2} + B x + C) = 2 x^{2} + 3 x \\ 2 A - 6 A x - 3 B + A x^{2} + B x + C = 2 x^{2} + 3 x \\ \underset{1}{\underset{―}{A}} x^{2} + \underset{2}{\underset{―}{(- 6 A + B)}} x + \underset{3}{\underset{―}{(2 A - 3 B + C)}} = \underset{1}{\underset{―}{2}} x^{2} + \underset{2}{\underset{―}{3}} x + \underset{3}{\underset{―}{0}} . \end{matrix}

Matching the underlined coefficients leads to the following system of equations and solution for the coefficients

A,

B,

and

C :

\begin{aligned} 1. \\ 2. \\ 3. \end{aligned}

\begin{aligned} A = & 2 \\ - 6 A + B = & 3 \\ 2 A - 3 B + C = & 0 \end{aligned}

\begin{matrix} \to \end{matrix}

\begin{aligned} A = 2 \\ B = 9 \\ C = 13 \end{aligned}

The general solution is then

y = y_{h} + y_{p} = c_{1} e^{x} + c_{2} x e^{x} + 2 x^{2} + 9 x + 13 .

Solution 2.

y^{″} + 3 y^{'} - 28 y = 7 t + e^{4 t} - 1

Following the method of Undetermined Coefficients, we find

y_{h}

y_{h} = c_{1} e^{4 t} + c_{2} e^{- 7 t}

and then find

y_{p}

through the steps that follow.

Select Initial

y_{p}

. The initial form of

y_{p}

y_{p} = A t + B + C e^{4 t},

Adjust

y_{p}

. Notice that

y_{p}

has an

e^{4 t}

term, which overlaps with

y_{h} .

Therefore, we need to adjust

y_{p}

by multiplying the

e^{4 t}

term by

t,

y_{p} = A t + B + C t e^{4 t} .

Now that

y_{p}

and

y_{h}

are independent, we can proceed to find the coefficients

A,

B,

and

C .

Find the Coefficients of

y_{p}

. Before plugging in

y_{p},

let’s find

y_{p}^{'}

and

y_{p}^{″} :

\begin{aligned} y_{p} = & A t + B + C t e^{4 t} \\ y_{p}^{'} = & A + C e^{4 t} + 4 C t e^{4 t} \\ y_{p}^{″} = & 8 C e^{4 t} + 16 C t e^{4 t} . \end{aligned}

Now, substituting these into the equation and collecting like terms, to get

\begin{aligned} 8 C e^{4 t} + 16 C t e^{4 t} & + 3 (A + C e^{4 t} + 4 C t e^{4 t}) \\ - 28 (A t + B + C t e^{4 t}) = 7 t + e^{4 t} - 1 \\ \underset{1}{\underset{―}{(11 C)}} e^{t} + \underset{2}{\underset{―}{(- 28 A)}} & t + \underset{3}{\underset{―}{(3 A - 28 B)}} = \underset{1}{\underset{―}{(1)}} e^{4 t} + \underset{2}{\underset{―}{(7)}} t + \underset{3}{\underset{―}{(- 1)}}, \end{aligned}

which leads to the following equations for

A,

B,

and

C :

\begin{aligned} 1. \\ 2. \\ 3. \end{aligned}

\begin{aligned} 11 C = & 1 \\ - 28 A = & 7 \\ 3 A - 28 B = & - 1 \end{aligned}

\begin{matrix} \to \end{matrix}

\begin{aligned} C = 1 / 11 \\ A = - 1 / 4 \\ B = 1 / 112 \end{aligned}

Finally, the general solution is

y = y_{h} + y_{p} = c_{1} e^{4 t} + c_{2} e^{- 7 t} - \frac{1}{4} t + \frac{1}{112} + \frac{1}{11} e^{4 t} .

Reading Questions Check-Point Questions

1. After substituting $y_{p} = A x + B$ into the equation $y^{″} - 4 y^{'} + 3 y = 9 x,$ you find the system of equations...

y_{p} = A x + B

y^{″} - 4 y^{'} + 3 y = 9 x,

3 A = 9

- 4 A + 3 B = 0 .

A

B ?

$A = 3, B = 4$
Correct! Solving the system yields $A = 3$ and $B = 4 .$
$A = 1, B = 2$
Incorrect. Double-check your algebra when solving the system of equations.
$A = 3, B = 0$
Incorrect. While $A$ is correct, you need to solve for $B$ as well.
$A = 2, B = 4$
Incorrect. $A = 3,$ not 2.

2. Which of the following is NOT a step in finding the coefficients of the particular solution?

Which of the following is NOT a step in finding the coefficients of the particular solution?

Plugging $y_{p}$ into the differential equation.
Incorrect. This is the first step in determining the coefficients.
Applying the initial conditions.
Correct! The initial conditions are used to solve for the unknown constants in the general solution, not the particular solution.
Collecting and matching like terms.
Incorrect. This is a crucial step for finding the values of $A$ and $B .$
Solving the resulting system of equations.
Incorrect. This is the final step in finding the coefficients.

3. Consider the equation...

\begin{matrix} (42) & y^{″} - 4 y^{'} + 3 y = 9 x . \end{matrix}

y_{p} = A x + B,

$3 A = 9, - 4 A + 3 B = 9$
Incorrect. Remember to match coefficients of like terms separately (i.e., coefficients of $x$ and constant terms).
$3 A = 9, - 4 A + 3 B = 0$
Correct! Matching coefficients of $x$ gives $3 A = 9,$ and matching constant terms gives $- 4 A + 3 B = 0 .$
$A = 9, - 4 A + B = 0$
Incorrect. Re-check the coefficients and ensure you match the terms correctly.
$3 A = 3, - 4 B + A = 9$
Incorrect. Verify which terms correspond to coefficients of $x$ and the constant terms.

4. Why do we adjust the form of $y_{p}$ by multiplying terms by $x$ when there is overlap with $y_{h} ?$

Why do we adjust the form of

y_{p}

by multiplying terms by

x

when there is overlap with

y_{h} ?

To make the computation of derivatives easier.
Incorrect. The adjustment is not made for the ease of differentiation but to ensure the independence of terms.
To ensure that $y_{p}$ and $y_{h}$ remain independent.
Correct! Multiplying by $x$ ensures that the particular solution does not overlap with the homogeneous solution.
To match the degree of the polynomial in $f (x) .$
Incorrect. While the degree of $y_{p}$ is related to $f (x),$ the adjustment addresses overlap with $y_{h} .$
To simplify the resulting system of equations.
Incorrect. The adjustment is made to maintain independence between $y_{p}$ and $y_{h},$ not for simplification purposes.

5. Suppose
$y^{″} - 4 y^{'} + 3 y = 6 e^{x} .$
Which of the following is the correct initial guess for $y_{p} ?$

y^{″} - 4 y^{'} + 3 y = 6 e^{x} .

y_{p} ?

$y_{p} = A e^{x}$
Incorrect. Remember to consider whether $e^{x}$ overlaps with any terms in $y_{h} .$
$y_{p} = A x e^{x}$
Correct! Since $e^{x}$ is a term in $y_{h},$ we need to multiply the particular solution by $x$ to ensure independence.
$y_{p} = A x + B$
Incorrect. This form is more suited for a polynomial right-hand side like $f (x) = 6 x .$
$y_{p} = A$
Incorrect. A constant form of $y_{p}$ would only be appropriate if $f (x)$ were a constant.

6. What are the main steps for finding $y_{p}$ using the method of undetermined coefficients?

What are the main steps for finding

y_{p}

using the method of undetermined coefficients?

Select the particular solution, integrate it, and solve for coefficients.
Incorrect. Integration is not a part of the method of undetermined coefficients.
Select the particular solution, adjust it if necessary, and find its coefficients.
Correct! The process involves making an educated guess, substituting it, and solving for unknown coefficients.
Select the homogeneous solution, adjust for $y_{p},$ and differentiate.
Incorrect. The focus is on selecting and solving for $y_{p},$ not guessing the homogeneous solution.
Choose $y_{p},$ integrate the result, and verify the solution.
Incorrect. Integration is not a step in this method; it’s about solving for coefficients of $y_{p} .$

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