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Mathematics LibreTexts

2.4: The Method of Undetermined Coefficient

  • Page ID
    17151
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    In this section we consider the constant coefficient equation

    \begin{equation}\label{eq:2.4.1}
    ay''+by'+cy=e^{\alpha x}G(x),
    \end{equation}

    where \(\alpha\) is a constant and \(G\) is a polynomial.

    From Theorem \((2.3.2)\), the general solution of (\ref{eq:2.4.1}) is \(y=y_p+c_1y_1+c_2y_2\), where \(y_p\) is a particular solution of (\ref{eq:2.4.1}) and \(\{y_1,y_2\}\) is a fundamental set of solutions of the complementary equation

    \begin{eqnarray*}
    ay''+by'+cy=0.
    \end{eqnarray*}

    In Section \(2.2\) 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 \( \textcolor{blue}{\mbox{the method of undetermined coefficients}} \).

    Our first example is similar to Exercises \((2.3E.16)\) to \((2.3E.21)\).

    Example \(\PageIndex{1}\)

    Find a particular solution of

    \begin{equation}\label{eq:2.4.2}
    y''-7y'+12y=4e^{2x}.
    \end{equation}

    Then find the general solution.

     

    Answer

    Substituting \(y_p=Ae^{2x}\) for \(y\) in (\ref{eq:2.4.2}) will produce a constant multiple of \(Ae^{2x}\) on the left side of (\ref{eq:2.4.2}), so it may be possible to choose \(A\) so that \(y_p\) is a solution of (\ref{eq:2.4.2}). Let's try it; if \(y_p=Ae^{2x}\) then

    \begin{eqnarray*}
    y_p''-7y_p'+12y_p=4Ae^{2x}-14Ae^{2x}+12Ae^{2x}=2Ae^{2x}=4e^{2x}
    \end{eqnarray*}

    if \(A=2\). Therefore \(y_p=2e^{2x}\) is a particular solution of (\ref{eq:2.4.2}). To find the general solution, we note that the characteristic polynomial of the complementary equation 

    \begin{equation}\label{eq:2.4.3}
    y''-7y'+12y=0
    \end{equation}

    is \(p(r)=r^2-7r+12=(r-3)(r-4)\), so \(\{e^{3x},e^{4x}\}\) is a fundamental set of solutions of (\ref{eq:2.4.3}). Therefore the general solution of (\ref{eq:2.4.2}) is

    \begin{eqnarray*}
    y=2e^{2x}+c_1e^{3x}+c_2e^{4x}.
    \end{eqnarray*}

    Example \(\PageIndex{2}\)

    Find a particular solution of

    \begin{equation}\label{eq:2.4.4}
    y''-7y'+12y=5e^{4x}.
    \end{equation}

    Then find the general solution.

     

    Answer

    Fresh from our success in finding a particular solution of \ref{eq:2.4.2} (where we chose \(y_p=Ae^{2x}\) because the right side of \ref{eq:2.4.2} is a constant multiple of \(e^{2x}\)) it may seem reasonable to try \(y_p=Ae^{4x}\) as a particular solution of \ref{eq:2.4.4}. However, this won't work, since we saw in Example \((2.4.1)\) that \(e^{4x}\) is a solution of the complementary equation \ref{eq:2.4.3}, so substituting \(y_p=Ae^{4x}\) into the left side of \ref{eq:2.4.4} 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 \(2.2\) to find a second solution of

    \begin{eqnarray*}
    ay''+by'+cy=0
    \end{eqnarray*}

    in the case where the characteristic equation has a repeated real root: we look for solutions of \ref{eq:2.4.4} in the form \(y=ue^{4x}\), where \(u\) is a function to be determined. Substituting

    \begin{equation}\label{eq:2.4.5}
    y=ue^{4x},\quad y'=u'e^{4x}+4ue^{4x}, \quad \mbox{and} \quad y''=u''e^{4x}+8u'e^{4x}+16ue^{4x}
    \end{equation}

    into \ref{eq:2.4.4} and canceling the common factor \(e^{4x}\) yields

    \begin{eqnarray*}
    (u''+8u'+16u)-7(u'+4u)+12u=5,
    \end{eqnarray*}

    or

    \begin{eqnarray*}
    u''+u'=5.
    \end{eqnarray*}

    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 \ref{eq:2.4.4}. Therefore

    \begin{eqnarray*}
    y=5xe^{4x}+c_1e^{3x}+c_2e^{4x}
    \end{eqnarray*}

    is the general solution.

    Example \(\PageIndex{3}\)

    Find a particular solution of

    \begin{equation}\label{eq:2.4.6}
    y''-8y'+16y=2e^{4x}.
    \end{equation}

     

    Answer

    Since the characteristic polynomial of the complementary equation

    \begin{equation}\label{eq:2.4.7}
    y''-8y'+16y=0
    \end{equation}

    is \(p(r)=r^2-8r+16=(r-4)^2\), both \(y_1=e^{4x}\) and \(y_2=xe^{4x}\) are solutions of \ref{eq:2.4.7}. Therefore \ref{eq:2.4.6) does not have a solution of the form \(y_p=Ae^{4x}\) or \(y_p=Axe^{4x}\). As in Example \(2.4.2)\), we look for solutions of \ref{eq:2.4.6} in the form \(y=ue^{4x}\), where \(u\) is a function to be determined. Substituting from \ref{eq:2.4.5} into \ref{eq:2.4.6} and canceling the common factor \(e^{4x}\) yields

    \begin{eqnarray*}
    (u''+8u'+16u)-8(u'+4u)+16u=2,
    \end{eqnarray*}

    or

    \begin{eqnarray*}
    u''=2.
    \end{eqnarray*}

    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 \ref{eq:2.4.4}. Therefore

    \begin{eqnarray*}
    y=e^{4x}(x^2+c_1+c_2x)
    \end{eqnarray*}

    is the general solution.

    The preceding examples illustrate the following facts concerning the form of a particular solution \(y_p\) of a constant coefficient equation

    \begin{eqnarray*}
    ay''+by'+cy=ke^{\alpha x},
    \end{eqnarray*}

    where \(k\) is a nonzero constant:

    (a) If \(e^{\alpha x}\) isn't a solution of the complementary equation

    \begin{equation}\label{eq:2.4.8}
    ay''+by'+cy=0,
    \end{equation}

    then \(y_p=Ae^{\alpha x}\), where \(A\) is a constant. (See Example \((2.4.1)\).)

     

    (b) If \(e^{\alpha x}\) is a solution of \ref{eq:2.4.8} but \(xe^{\alpha x}\) is not, then \(y_p=Axe^{\alpha x}\), where \(A\) is a constant. (See Example \((2.4.2)\).)

    (c) If both \(e^{\alpha x}\) and \(xe^{\alpha x}\) are solutions of \ref{eq:2.4.8}, then \(y_p=Ax^2e^{\alpha x}\), where \(A\) is a constant. (See Example \((2.4.3)\).)

    See Exercise \((2.4E.30)\) 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

    \begin{eqnarray*}
    ay_p''+by_p'+cy_p=ke^{\alpha x},
    \end{eqnarray*}

    and solve for the constant \(A\), as we did in Example \((2.4.1)\). (See Exercises \((2.4E.31)\) to \((2.4E.33)\).) However, if the equation is

    \begin{eqnarray*}
    ay''+by'+cy=k e^{\alpha x}G(x),
    \end{eqnarray*}

    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 \((2.4.2)\) and \((2.4.3)\). The equation for \(u\) will turn out to be

    \begin{equation}\label{eq:2.4.9}
    au''+p'(\alpha)u'+p(\alpha)u=G(x),
    \end{equation}

    where \(p(r)=ar^2+br+c\) is the characteristic polynomial of the complementary equation and \(p'(r)=2ar+b\) (Exercise \((2.4E.30)\)); however, you shouldn't memorize this since it's 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 \ref{eq:2.4.9} reduces to

    \begin{eqnarray*}
    au''+p'(\alpha)u'=G(x),
    \end{eqnarray*}

    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 \ref{eq:2.4.9} reduces to

    \begin{eqnarray*}
    au''=G(x).
    \end{eqnarray*}

    Example \(\PageIndex{4}\)

    Find a particular solution of

    \begin{equation}\label{eq:2.4.10}
    y''-3y'+2y=e^{3x}(-1+2x+x^2).
    \end{equation}

     

    Answer

    Substituting

    \begin{eqnarray*}
    y=ue^{3x},\quad y'=u'e^{3x}+3ue^{3x}, \quad \mbox{and} \quad y''=u''e^{3x}+6u'e^{3x}+9ue^{3x}
    \end{eqnarray*}

    into \ref{eq:2.4.10} and canceling \(e^{3x}\) yields

    \begin{eqnarray*}
    (u''+6u'+9u)-3(u'+3u)+2u=-1+2x+x^2,
    \end{eqnarray*}

    or

    \begin{equation}\label{eq:2.4.11}
    u''+3u'+2u=-1+2x+x^2.
    \end{equation}

    As in Example \((2.3.2)\), in order to guess a form for a particular solution of \ref{eq:2.4.11}, we note that substituting a second degree polynomial \(u_p=A+Bx+Cx^2\) for \(u\) in the left side of \ref{eq:2.4.11} produces another second degree polynomial with coefficients that depend upon \(A\), \(B\), and \(C\); thus, 

    \begin{eqnarray*}
    \mbox{ if} \quad u_p=A+Bx+Cx^2 \quad \mbox{then} \quad u_p'=B+2Cx \quad  \mbox{and} \quad u_p''=2C.
    \end{eqnarray*}

    If \(u_p\) is to satisfy \ref{eq:2.4.11}, we must have 

    \begin{eqnarray*}
    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{eqnarray*}

    Equating coefficients of like powers of \(x\) on the two sides of the last equality yields

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

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

    \begin{eqnarray*}
    u_p=-{1\over4}(1+2x-2x^2)
    \end{eqnarray*}

    is a particular solution of \ref{eq:2.4.11}, and

    \begin{eqnarray*}
    y_p=u_pe^{3x}=-{e^{3x}\over4}(1+2x-2x^2)
    \end{eqnarray*}

    is a particular solution of \ref{eq:2.4.10}.

    Example \(\PageIndex{5}\)

    Find a particular solution of

    \begin{equation}\label{eq:2.4.12}
    y''-4y'+3y=e^{3x}(6+8x+12x^2).
    \end{equation}

     

    Answer

    Substituting

    \begin{eqnarray*}
    y=ue^{3x},\quad y'=u'e^{3x}+3ue^{3x},\quad \mbox{and} \quad y''=u''e^{3x}+6u'e^{3x}+9ue^{3x}
    \end{eqnarray*}

    into \ref{eq:2.4.12} and canceling \(e^{3x}\) yields

    \begin{eqnarray*}
    (u''+6u'+9u)-4(u'+3u)+3u=6+8x+12x^2,
    \end{eqnarray*}

    or

    \begin{equation}\label{eq:2.4.13}
    u''+2u'=6+8x+12x^2.
    \end{equation}

    There's no \(u\) term in this equation, since \(e^{3x}\) is a solution of the complementary equation for \ref{eq:2.4.12}. (See

    Exercise \((2.4E.30)\).) Therefore \ref{eq:2.4.13} does not have a particular solution of the form \(u_p=A+Bx+Cx^2\) that we used successfully in Example \((2.4.4)\), since with this choice of \(u_p\),

    \begin{eqnarray*}
    u_p''+2u_p'=2C+(B+2Cx)
    \end{eqnarray*}

    can't contain the last term (\(12x^2\)) on the right side of \ref{eq:2.4.13}. Instead, let's try \(u_p=Ax+Bx^2+Cx^3\) on the grounds that

    \begin{eqnarray*}
    u_p'=A+2Bx+3Cx^2 \quad \mbox{and} \quad u_p''=2B+6Cx
    \end{eqnarray*}

    together contain all the powers of \(x\) that appear on the right side of \ref{eq:2.4.13}.

    Substituting these expressions in place of \(u'\) and \(u''\) in \ref{eq:2.4.13} yields

    \begin{eqnarray*}
    (2B+6Cx)+2(A+2Bx+3Cx^2)=(2B+2A)+(6C+4B)x+6Cx^2=6+8x+12x^2.
    \end{eqnarray*}

    Comparing coefficients of like powers of \(x\) on the two sides of the last equality shows that \(u_p\) satisfies \ref{eq:2.4.13} if

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

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

    \begin{eqnarray*}
    u_p=x(4-x+2x^2)
    \end{eqnarray*}

    is a particular solution of \ref{eq:2.4.13}, and

    \begin{eqnarray*}
    y_p=u_pe^{3x}=xe^{3x}(4-x+2x^2)
    \end{eqnarray*}

    is a particular solution of \ref{eq:2.4.12}.

    Example \(\PageIndex{6}\)

    Find a particular solution of

    \begin{equation}\label{eq:2.4.14}
    4y''+4y'+y=e^{-x/2}(-8+48x+144x^2).
    \end{equation}

     

    Answer

    Substituting

    \begin{eqnarray*}
    y=ue^{-x/2},\quad y'=u'e^{-x/2}-{1\over2}ue^{-x/2}, \quad \mbox{and} \quad y''=u''e^{-x/2}-u'e^{-x/2}+{1\over4}ue^{-x/2}
    \end{eqnarray*}

    into \ref{eq:2.4.14} and canceling \(e^{-x/2}\) yields

    \begin{eqnarray*}
    4\left(u''-u'+{u\over4}\right)+4\left(u'-{u\over2}\right)+u=4u''=-8+48x+144x^2,
    \end{eqnarray*}

    or

    \begin{equation}\label{eq:2.4.15}
    u''=-2+12x+36x^2,
    \end{equation}

    which does not contain \(u\) or \(u'\) because \(e^{-x/2}\) and \(xe^{-x/2}\) are both solutions of the complementary equation. (See Exercise \((2.4E.30)\).) To obtain a particular solution of \ref{eq:2.4.15} we integrate twice, taking the constants of integration to be zero; thus,

    \begin{eqnarray*}
    u_p'=-2x+6x^2+12x^3 \quad \mbox{and} \quad u_p=-x^2+2x^3+3x^4=x^2(-1+2x+3x^2).
    \end{eqnarray*}

    Therefore

    \begin{eqnarray*}
    y_p=u_pe^{-x/2}=x^2e^{-x/2}(-1+2x+3x^2)
    \end{eqnarray*}

    is a particular solution of \ref{eq:2.4.14}. 

     

    Summary

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

    \begin{eqnarray*}
    ay''+by'+cy=e^{\alpha x}G(x),
    \end{eqnarray*}

    where \(G\) is a polynomial (see Exercise \((2.4E.30)\)):

    (a) If \(e^{\alpha x}\) isn't a solution of the complementary equation

    \begin{equation}\label{eq:2.4.16}
    ay''+by'+cy=0,
    \end{equation}

    then \(y_p=e^{\alpha x}Q(x)\), where \(Q\) is a polynomial of the same degree as \(G\). (See Example \((2.4.4)\)).

    (b) If \(e^{\alpha x}\) is a solution of \ref{eq:2.4.16} 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 \((2.4.5)\).)

    (c) If both \(e^{\alpha x}\) and \(xe^{\alpha x}\) are solutions of \ref{eq:2.4.16}, then \(y_p=x^2e^{\alpha x}Q(x)\), where \(Q\) is a polynomial of the same degree as \(G\). (See Example \((2.4.6)\).)

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

    \begin{eqnarray*}
    ay_p''+by_p'+cy_p=e^{\alpha x}G(x),
    \end{eqnarray*}

    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 \((2.4E.34)\) to \((2.4E.36)\).) In Case (a) the equation for \(u\) will be of the form 

    \begin{eqnarray*}
    au''+p'(\alpha)u'+p(\alpha)u=G(x),
    \end{eqnarray*}

    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 \((2.4.4)\). In Case (b) the equation for \(u\) will be of the form

    \begin{eqnarray*}
    au''+p'(\alpha)u'=G(x)
    \end{eqnarray*}

    (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 \((2.4.5)\). In Case (c) the equation for \(u\) will be of the form

    \begin{eqnarray*}
    au''=G(x)
    \end{eqnarray*}

    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 \((2.4.6)\).

    Using the Principle of Superposition

    The next example shows how to combine the method of undetermined coefficients and Theorem \((2.3.3)\), the principle of

    superposition.

    Example \(\PageIndex{7}\)

    Find a particular solution of

    \begin{equation}\label{eq:2.4.17}
    y''-7y'+12y=4e^{2x}+5e^{4x}.
    \end{equation}

     

    Answer

    In Example \((2.4.1)\) we found that \(y_{p_1}=2e^{2x}\) is a particular solution of

    \begin{eqnarray*}
    y''-7y'+12y=4e^{2x},
    \end{eqnarray*}

    and in Example \((2.4.2)\) we found that \(y_{p_2}=5xe^{4x}\) is a particular solution of

    \begin{eqnarray*}
    y''-7y'+12y=5e^{4x}.
    \end{eqnarray*}

    Therefore the principle of superposition implies that \(y_p=2e^{2x}+5xe^{4x}\) is a particular solution of \ref{eq:2.4.17}.