5.4: The Method of Undetermined Coefficients I


In this section we consider the constant coefficient equation

$\label{eq:5.4.1} ay''+by'+cy=e^{\alpha x}G(x),$

where $$\alpha$$ is a constant and $$G$$ is a polynomial.

From Theorem 5.3.2, the general solution of Equation \ref{eq:5.4.1} is $$y=y_p+c_1y_1+c_2y_2$$, where $$y_p$$ is a particular solution of Equation \ref{eq:5.4.1} and $$\{y_1,y_2\}$$ is a fundamental set of solutions of the complementary equation

$ay''+by'+cy=0. \nonumber$

In Section 5.2 we showed how to find $$\{y_1,y_2\}$$. In this section we’ll show how to find $$y_p$$. The procedure that we’ll use is called the method of undetermined coefficients. Our first example is similar to Exercises 5.3.16-5.3.21.

Example 5.4.1

Find a particular solution of

$\label{eq:5.4.2} y''-7y'+12y=4e^{2x}.$

Then find the general solution.

Solution

Substituting $$y_p=Ae^{2x}$$ for $$y$$ in Equation \ref{eq:5.4.2} will produce a constant multiple of $$Ae^{2x}$$ on the left side of Equation \ref{eq:5.4.2}, so it may be possible to choose $$A$$ so that $$y_p$$ is a solution of Equation \ref{eq:5.4.2}. Let’s try it; if $$y_p=Ae^{2x}$$ then

$y_p''-7y_p'+12y_p=4Ae^{2x}-14Ae^{2x}+12Ae^{2x}=2Ae^{2x}=4e^{2x} \nonumber$

if $$A=2$$. Therefore $$y_p=2e^{2x}$$ is a particular solution of Equation \ref{eq:5.4.2}. To find the general solution, we note that the characteristic polynomial of the complementary equation

$\label{eq:5.4.3} y''-7y'+12y=0$

is $$p(r)=r^2-7r+12=(r-3)(r-4)$$, so $$\{e^{3x},e^{4x}\}$$ is a fundamental set of solutions of Equation \ref{eq:5.4.3}. Therefore the general solution of Equation \ref{eq:5.4.2} is

$y=2e^{2x}+c_1e^{3x}+c_2e^{4x}. \nonumber$

Example 5.4.2

Find a particular solution of

$\label{eq:5.4.4} y''-7y'+12y=5e^{4x}.$

Then find the general solution.

Solution

Fresh from our success in finding a particular solution of Equation \ref{eq:5.4.2} — where we chose $$y_p=Ae^{2x}$$ because the right side of Equation \ref{eq:5.4.2} is a constant multiple of $$e^{2x}$$ — it may seem reasonable to try $$y_p=Ae^{4x}$$ as a particular solution of Equation \ref{eq:5.4.4}. However, this will not work, since we saw in Example 5.4.1 that $$e^{4x}$$ is a solution of the complementary equation Equation \ref{eq:5.4.3}, so substituting $$y_p=Ae^{4x}$$ into the left side of Equation \ref{eq:5.4.4}) produces zero on the left, no matter how we choose$$A$$. To discover a suitable form for $$y_p$$, we use the same approach that we used in Section 5.2 to find a second solution of

$ay''+by'+cy=0 \nonumber$

in the case where the characteristic equation has a repeated real root: we look for solutions of Equation \ref{eq:5.4.4} in the form $$y=ue^{4x}$$, where $$u$$ is a function to be determined. Substituting

$\label{eq:5.4.5} y=ue^{4x},\quad y'=u'e^{4x}+4ue^{4x},\quad \text{and} \quad y''=u''e^{4x}+8u'e^{4x}+16ue^{4x}$

into Equation \ref{eq:5.4.4} and canceling the common factor $$e^{4x}$$ yields

$(u''+8u'+16u)-7(u'+4u)+12u=5, \nonumber$

or

$u''+u'=5. \nonumber$

By inspection we see that $$u_p=5x$$ is a particular solution of this equation, so $$y_p=5xe^{4x}$$ is a particular solution of Equation \ref{eq:5.4.4}. Therefore

$y=5xe^{4x}+c_1e^{3x}+c_2e^{4x} \nonumber$

is the general solution.

Example 5.4.3

Find a particular solution of

$\label{eq:5.4.6} y''-8y'+16y=2e^{4x}.$

Solution

Since the characteristic polynomial of the complementary equation

$\label{eq:5.4.7} y''-8y'+16y=0$

is $$p(r)=r^2-8r+16=(r-4)^2$$, both $$y_1=e^{4x}$$ and $$y_2=xe^{4x}$$ are solutions of Equation \ref{eq:5.4.7}. Therefore Equation \ref{eq:5.4.6}) does not have a solution of the form $$y_p=Ae^{4x}$$ or $$y_p=Axe^{4x}$$. As in Example 5.4.2 , we look for solutions of Equation \ref{eq:5.4.6} in the form $$y=ue^{4x}$$, where $$u$$ is a function to be determined. Substituting from Equation \ref{eq:5.4.5} into Equation \ref{eq:5.4.6} and canceling the common factor $$e^{4x}$$ yields

$(u''+8u'+16u)-8(u'+4u)+16u=2, \nonumber$

or

$u''=2. \nonumber$

Integrating twice and taking the constants of integration to be zero shows that $$u_p=x^2$$ is a particular solution of this equation, so $$y_p=x^2e^{4x}$$ is a particular solution of Equation \ref{eq:5.4.4}. Therefore

$y=e^{4x}(x^2+c_1+c_2x) \nonumber$

is the general solution.

The preceding examples illustrate the following facts concerning the form of a particular solution $$y_p$$ of a constant coefficent equation

$ay''+by'+cy=ke^{\alpha x}, \nonumber$

where $$k$$ is a nonzero constant:

1. If $$e^{\alpha x}$$ isn’t a solution of the complementary equation $\label{eq:5.4.8} ay''+by'+cy=0,$ then $$y_p=Ae^{\alpha x}$$, where $$A$$ is a constant. (See Example 5.4.1 ).
2. If $$e^{\alpha x}$$ is a solution of Equation \ref{eq:5.4.8} but $$xe^{\alpha x}$$ is not, then $$y_p=Axe^{\alpha x}$$, where $$A$$ is a constant. (See Example 5.4.2 .)
3. If both $$e^{\alpha x}$$ and $$xe^{\alpha x}$$ are solutions of Equation \ref{eq:5.4.8}, then $$y_p=Ax^2e^{\alpha x}$$, where $$A$$ is a constant. (See Example 5.4.3 .)

See Exercise 5.4.30 for the proofs of these facts.

In all three cases you can just substitute the appropriate form for $$y_p$$ and its derivatives directly into

$ay_p''+by_p'+cy_p=ke^{\alpha x},\nonumber$

and solve for the constant $$A$$, as we did in Example 5.4.1 . (See Exercises 5.4.31-5.4.33.) However, if the equation is

$ay''+by'+cy=k e^{\alpha x}G(x), \nonumber$

where $$G$$ is a polynomial of degree greater than zero, we recommend that you use the substitution $$y=ue^{\alpha x}$$ as we did in Examples 5.4.2 and 5.4.3 . The equation for $$u$$ will turn out to be

$\label{eq:5.4.9} au''+p'(\alpha)u'+p(\alpha)u=G(x),$

where $$p(r)=ar^2+br+c$$ is the characteristic polynomial of the complementary equation and $$p'(r)=2ar+b$$ (Exercise 5.4.30); however, you shouldn’t memorize this since it is easy to derive the equation for $$u$$ in any particular case. Note, however, that if $$e^{\alpha x}$$ is a solution of the complementary equation then $$p(\alpha)=0$$, so Equation \ref{eq:5.4.9} reduces to

$au''+p'(\alpha)u'=G(x), \nonumber$

while if both $$e^{\alpha x}$$ and $$xe^{\alpha x}$$ are solutions of the complementary equation then $$p(r)=a(r-\alpha)^2$$ and $$p'(r)=2a(r-\alpha)$$, so $$p(\alpha)=p'(\alpha)=0$$ and Equation \ref{eq:5.4.9}) reduces to

$au''=G(x). \nonumber$

Example 5.4.4

Find a particular solution of

$\label{eq:5.4.10} y''-3y'+2y=e^{3x}(-1+2x+x^2).$

Solution

Substituting

$y=ue^{3x},\quad y'=u'e^{3x}+3ue^{3x},\quad \text{and} y''=u''e^{3x}+6u'e^{3x}+9ue^{3x}\nonumber$

into Equation \ref{eq:5.4.10}) and canceling $$e^{3x}$$ yields

$(u''+6u'+9u)-3(u'+3u)+2u=-1+2x+x^2, \nonumber$

or

$\label{eq:5.4.11} u''+3u'+2u=-1+2x+x^2.$

As in Example 5.3.2, in order to guess a form for a particular solution of Equation \ref{eq:5.4.11}), we note that substituting a second degree polynomial $$u_p=A+Bx+Cx^2$$ for $$u$$ in the left side of Equation \ref{eq:5.4.11}) produces another second degree polynomial with coefficients that depend upon $$A$$, $$B$$, and $$C$$; thus,

$\text{if} \quad u_p=A+Bx+Cx^2\quad \text{then} \quad u_p'=B+2Cx\quad \text{and} \quad u_p''=2C. \nonumber$

If $$u_p$$ is to satisfy Equation \ref{eq:5.4.11}), we must have

\begin{aligned} u_p''+3u_p'+2u_p&=2C+3(B+2Cx)+2(A+Bx+Cx^2)\\ &=(2C+3B+2A)+(6C+2B)x+2Cx^2=-1+2x+x^2.\end{aligned}\nonumber

Equating coefficients of like powers of $$x$$ on the two sides of the last equality yields

$\begin{array}{rcr} 2C&=1\phantom{.}\\ 2B+6C&=2\phantom{.}\\ 2A+3B+2C&= -1. \end{array}\nonumber$

Solving these equations for $$C$$, $$B$$, and $$A$$ (in that order) yields $$C=1/2,B=-1/2,A=-1/4$$. Therefore

$u_p=-{1\over4}(1+2x-2x^2) \nonumber$

is a particular solution of Equation \ref{eq:5.4.11}, and

$y_p=u_pe^{3x}=-{e^{3x}\over4}(1+2x-2x^2) \nonumber$

is a particular solution of Equation \ref{eq:5.4.10}.

Example 5.4.5

Find a particular solution of

$\label{eq:5.4.12} y''-4y'+3y=e^{3x}(6+8x+12x^2).$

Solution

Substituting

$y=ue^{3x},\quad y'=u'e^{3x}+3ue^{3x},\quad \text{and } y''=u''e^{3x}+6u'e^{3x}+9ue^{3x} \nonumber$

into Equation \ref{eq:5.4.12}) and canceling $$e^{3x}$$ yields

$(u''+6u'+9u)-4(u'+3u)+3u=6+8x+12x^2, \nonumber$

or

$\label{eq:5.4.13} u''+2u'=6+8x+12x^2.$

There’s no $$u$$ term in this equation, since $$e^{3x}$$ is a solution of the complementary equation for Equation \ref{eq:5.4.12}). (See Exercise 5.4.30.) Therefore Equation \ref{eq:5.4.13}) does not have a particular solution of the form $$u_p=A+Bx+Cx^2$$ that we used successfully in Example 5.4.4 , since with this choice of $$u_p$$,

$u_p''+2u_p'=2C+(B+2Cx) \nonumber$

can’t contain the last term ($$12x^2$$) on the right side of Equation \ref{eq:5.4.13}). Instead, let’s try $$u_p=Ax+Bx^2+Cx^3$$ on the grounds that

$u_p'=A+2Bx+3Cx^2\quad \text{and} \quad u_p''=2B+6Cx\nonumber$

together contain all the powers of $$x$$ that appear on the right side of Equation \ref{eq:5.4.13}).

Substituting these expressions in place of $$u'$$ and $$u''$$ in Equation \ref{eq:5.4.13}) yields

$(2B+6Cx)+2(A+2Bx+3Cx^2)=(2B+2A)+(6C+4B)x+6Cx^2=6+8x+12x^2. \nonumber$

Comparing coefficients of like powers of $$x$$ on the two sides of the last equality shows that $$u_p$$ satisfies Equation \ref{eq:5.4.13}) if

$\begin{array}{rcr} 6C&=12\phantom{.}\\ 4B+6C&=8\phantom{.}\\ 2A+2B\phantom{+6u_2}&=6. \end{array}\nonumber$

Solving these equations successively yields $$C=2$$, $$B=-1$$, and $$A=4$$. Therefore

$u_p=x(4-x+2x^2) \nonumber$

is a particular solution of Equation \ref{eq:5.4.13}), and

$y_p=u_pe^{3x}=xe^{3x}(4-x+2x^2) \nonumber$

is a particular solution of Equation \ref{eq:5.4.12}).

Example 5.4.6

Find a particular solution of

$\label{eq:5.4.14} 4y''+4y'+y=e^{-x/2}(-8+48x+144x^2).$

Solution

Substituting

$y=ue^{-x/2},\quad y'=u'e^{-x/2}-{1\over2}ue^{-x/2},\quad \text{and} \quad y''=u''e^{-x/2}-u'e^{-x/2}+{1\over4}ue^{-x/2} \nonumber$

into Equation \ref{eq:5.4.14}) and canceling $$e^{-x/2}$$ yields

$4\left(u''-u'+{u\over4}\right)+4\left(u'-{u\over2}\right)+u=4u''=-8+48x+144x^2, \nonumber$

or

$\label{eq:5.4.15} u''=-2+12x+36x^2,$

which does not contain $$u$$ or $$u'$$ because $$e^{-x/2}$$ and $$xe^{-x/2}$$ are both solutions of the complementary equation. (See Exercise 5.4.30.) To obtain a particular solution of Equation \ref{eq:5.4.15}) we integrate twice, taking the constants of integration to be zero; thus,

$u_p'=-2x+6x^2+12x^3\quad \text{and} \quad u_p=-x^2+2x^3+3x^4=x^2(-1+2x+3x^2).\nonumber$

Therefore

$y_p=u_pe^{-x/2}=x^2e^{-x/2}(-1+2x+3x^2)\nonumber$

is a particular solution of Equation \ref{eq:5.4.14}).

Summary

​​​​​The preceding examples illustrate the following facts concerning particular solutions of a constant coefficent equation of the form

$ay''+by'+cy=e^{\alpha x}G(x),\nonumber$

where $$G$$ is a polynomial (see Exercise 5.4.30):

1. If $$e^{\alpha x}$$ isn’t a solution of the complementary equation $\label{eq:5.4.16} ay''+by'+cy=0,$ then $$y_p=e^{\alpha x}Q(x)$$, where $$Q$$ is a polynomial of the same degree as $$G$$. (See Example 5.4.4 ).
2. If $$e^{\alpha x}$$ is a solution of Equation \ref{eq:5.4.16} but $$xe^{\alpha x}$$ is not, then $$y_p=xe^{\alpha x}Q(x)$$, where $$Q$$ is a polynomial of the same degree as $$G$$. (See Example 5.4.5 .)
3. If both $$e^{\alpha x}$$ and $$xe^{\alpha x}$$ are solutions of Equation \ref{eq:5.4.16}, then $$y_p=x^2e^{\alpha x}Q(x)$$, where $$Q$$ is a polynomial of the same degree as $$G$$. (See Example 5.4.6 .)

In all three cases, you can just substitute the appropriate form for $$y_p$$ and its derivatives directly into

$ay_p''+by_p'+cy_p=e^{\alpha x}G(x), \nonumber$

and solve for the coefficients of the polynomial $$Q$$. However, if you try this you will see that the computations are more tedious than those that you encounter by making the substitution $$y=ue^{\alpha x}$$ and finding a particular solution of the resulting equation for $$u$$. (See Exercises 5.4.34-5.4.36.) In Case (a) the equation for $$u$$ will be of the form

$au''+p'(\alpha)u'+p(\alpha)u=G(x), \nonumber$

with a particular solution of the form $$u_p=Q(x)$$, a polynomial of the same degree as $$G$$, whose coefficients can be found by the method used in Example 5.4.4 . In Case (b) the equation for $$u$$ will be of the form

$au''+p'(\alpha)u'=G(x) \nonumber$

(no $$u$$ term on the left), with a particular solution of the form $$u_p=xQ(x)$$, where $$Q$$ is a polynomial of the same degree as $$G$$ whose coefficents can be found by the method used in Example 5.4.5 . In Case (c), the equation for $$u$$ will be of the form

$au''=G(x) \nonumber$

with a particular solution of the form $$u_p=x^2Q(x)$$ that can be obtained by integrating $$G(x)/a$$ twice and taking the constants of integration to be zero, as in Example 5.4.6 .

Using the Principle of Superposition

The next example shows how to combine the method of undetermined coefficients and Theorem 5.3.3, the principle of superposition.

Example 5.4.7

Find a particular solution of

$\label{eq:5.4.17} y''-7y'+12y=4e^{2x}+5e^{4x}.$

Solution

In Example 5.4.1 we found that $$y_{p_1}=2e^{2x}$$ is a particular solution of

$y''-7y'+12y=4e^{2x}, \nonumber$

and in Example 5.4.2 we found that $$y_{p_2}=5xe^{4x}$$ is a particular solution of

$y''-7y'+12y=5e^{4x}. \nonumber$

Therefore the principle of superposition implies that $$y_p=2e^{2x}+5xe^{4x}$$ is a particular solution of Equation \ref{eq:5.4.17}).

This page titled 5.4: The Method of Undetermined Coefficients I is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.