# 5.4: The Method of Undetermined Coefficients I

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- Contributed by William F. Trench
- Andrew G. Cowles Distinguished Professor Emeritus (Mathamatics) at Trinity University

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 \(\PageIndex{1}\):

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 \(\PageIndex{2}\)

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 \(\PageIndex{1}\) 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 \(\PageIndex{3}\)

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 \(\PageIndex{2}\), 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:

- 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 \(\PageIndex{1}\)).
- 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 \(\PageIndex{2}\).)
- 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 \(\PageIndex{3}\).)

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 \(\PageIndex{1}\). (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 \(\PageIndex{2}\) and \(\PageIndex{3}\). 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 \(\PageIndex{4}\)

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 \(\PageIndex{5}\)

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 \(\PageIndex{4}\), 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 \(\PageIndex{6}\)

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*):

- 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 \(\PageIndex{4}\)).
- 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 \(\PageIndex{5}\).)
- 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 \(\PageIndex{6}\).)

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 \(\PageIndex{4}\). 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 \(\PageIndex{5}\). 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 \(\PageIndex{6}\).

## 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 \(\PageIndex{7}\)

Find a particular solution of

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

**Solution**

In Example \(\PageIndex{1}\) we found that \(y_{p_1}=2e^{2x}\) is a particular solution of

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

and in Example \(\PageIndex{2}\) 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}).