Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

5.6: Reduction of Order

  • Page ID
    30732
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    In this section we give a method for finding the general solution of

    \[\label{eq:5.6.1} P_0(x)y''+P_1(x)y'+P_2(x)y=F(x)\]

    if we know a nontrivial solution \(y_1\) of the complementary equation

    \[\label{eq:5.6.2} P_0(x)y''+P_1(x)y'+P_2(x)y=0.\]

    The method is called reduction of order because it reduces the task of solving Equation \ref{eq:5.6.1} to solving a first order equation. Unlike the method of undetermined coefficients, it does not require \(P_0\), \(P_1\), and \(P_2\) to be constants, or \(F\) to be of any special form.

    By now you shoudn’t be surprised that we look for solutions of Equation \ref{eq:5.6.1} in the form

    \[\label{eq:5.6.3} y=uy_1\]

    where \(u\) is to be determined so that \(y\) satisfies Equation \ref{eq:5.6.1}. Substituting Equation \ref{eq:5.6.3} and

    \[\begin{align*} y'&= u'y_1+uy_1' \\[4pt] y'' &= u''y_1+2u'y_1'+uy_1'' \end{align*}\]

    into Equation \ref{eq:5.6.1} yields

    \[P_0(x)(u''y_1+2u'y_1'+uy_1'')+P_1(x)(u'y_1+uy_1')+P_2(x)uy_1=F(x). \nonumber\]

    Collecting the coefficients of \(u\), \(u'\), and \(u''\) yields

    \[\label{eq:5.6.4} (P_0y_1)u''+(2P_0y_1'+P_1y_1)u'+(P_0y_1''+P_1y_1'+P_2y_1) u=F.\]

    However, the coefficient of \(u\) is zero, since \(y_1\) satisfies Equation \ref{eq:5.6.2}. Therefore Equation \ref{eq:5.6.4} reduces to

    \[\label{eq:5.6.5} Q_0(x)u''+Q_1(x)u'=F,\]

    with

    \[Q_0=P_0y_1 \quad \text{and} \quad Q_1=2P_0y_1'+P_1y_1.\nonumber\]

    (It isn’t worthwhile to memorize the formulas for \(Q_0\) and \(Q_1\)!) Since Equation \ref{eq:5.6.5} is a linear first order equation in \(u'\), we can solve it for \(u'\) by variation of parameters as in Section 1.2, integrate the solution to obtain \(u\), and then obtain \(y\) from Equation \ref{eq:5.6.3}.

    Example 5.6.1
    1. Find the general solution of \[\label{eq:5.6.6} xy''-(2x+1)y'+(x+1)y=x^2,\] given that \(y_1=e^x\) is a solution of the complementary equation \[\label{eq:5.6.7} xy''-(2x+1)y'+(x+1)y=0.\]
    2. As a byproduct of (a), find a fundamental set of solutions of Equation \ref{eq:5.6.7}.

    Solution

    a. If \(y=ue^x\), then \(y'=u'e^x+ue^x\) and \(y''=u''e^x+2u'e^x+ue^x\), so

    \[\begin{align*} xy''-(2x+1)y'+(x+1)y&=x(u''e^x+2u'e^x+ue^x) -(2x+1)(u'e^x+ue^x)+(x+1)ue^x\\ &=(xu''-u')e^x.\end{align*}\]

    Therefore \(y=ue^x\) is a solution of Equation \ref{eq:5.6.6} if and only if

    \[(xu''-u')e^x=x^2,\nonumber\]

    which is a first order equation in \(u'\). We rewrite it as

    \[\label{eq:5.6.8} u''-{u'\over x}=xe^{-x}.\]

    To focus on how we apply variation of parameters to this equation, we temporarily write \(z=u'\), so that Equation \ref{eq:5.6.8} becomes

    \[\label{eq:5.6.9} z'-{z\over x}=xe^{-x}.\]

    We leave it to you to show (by separation of variables) that \(z_1=x\) is a solution of the complementary equation

    \[z'-{z\over x}=0\nonumber\]

    for Equation \ref{eq:5.6.9}. By applying variation of parameters as in Section 1.2, we can now see that every solution of Equation \ref{eq:5.6.9} is of the form

    \[z=vx \quad \text{where} \quad v'x=xe^{-x}, \quad \text{so} \quad v'=e^{-x} \quad \text{and} \quad v=-e^{-x}+C_1.\nonumber\]

    Since \(u'=z=vx\), \(u\) is a solution of Equation \ref{eq:5.6.8} if and only if

    \[u'=vx=-xe^{-x}+C_1x.\nonumber\]

    Integrating this yields

    \[u=(x+1)e^{-x}+{C_1\over2}x^2+C_2.\nonumber\]

    Therefore the general solution of Equation \ref{eq:5.6.6} is

    \[\label{eq:5.6.10} y=ue^x=x+1+{C_1\over2}x^2e^x+C_2e^x.\]

    b. By letting \(C_1=C_2=0\) in Equation \ref{eq:5.6.10}, we see that \(y_{p_1}=x+1\) is a solution of Equation \ref{eq:5.6.6}. By letting \(C_1=2\) and \(C_2=0\), we see that \(y_{p_2}=x+1+x^2e^x\) is also a solution of Equation \ref{eq:5.6.6}. Since the difference of two solutions of Equation \ref{eq:5.6.6} is a solution of Equation \ref{eq:5.6.7}, \(y_2=y_{p_1}-y_{p_2}=x^2e^x\) is a solution of Equation \ref{eq:5.6.7}. Since \(y_2/y_1\) is nonconstant and we already know that \(y_1=e^x\) is a solution of Equation \ref{eq:5.6.6}, Theorem 5.1.6 implies that \(\{e^x,x^2e^x\}\) is a fundamental set of solutions of Equation \ref{eq:5.6.7}.

    Although Equation \ref{eq:5.6.10} is a correct form for the general solution of Equation \ref{eq:5.6.6}, it is silly to leave the arbitrary coefficient of \(x^2e^x\) as \(C_1/2\) where \(C_1\) is an arbitrary constant. Moreover, it is sensible to make the subscripts of the coefficients of \(y_1=e^x\) and \(y_2=x^2e^x\) consistent with the subscripts of the functions themselves. Therefore we rewrite Equation \ref{eq:5.6.10} as

    \[y=x+1+c_1e^x+c_2x^2e^x \nonumber\]

    by simply renaming the arbitrary constants. We’ll also do this in the next two examples, and in the answers to the exercises.

    Example 5.6.2
    1. Find the general solution of \[x^2y''+xy'-y=x^2+1, \nonumber\] given that \(y_1=x\) is a solution of the complementary equation \[\label{eq:5.6.11} x^2y''+xy'-y=0.\] As a byproduct of this result, find a fundamental set of solutions of Equation \ref{eq:5.6.11}.
    2. Solve the initial value problem \[\label{eq:5.6.12} x^2y''+xy'-y=x^2+1, \quad y(1)=2,\; y'(1)=-3.\]

    Solution

    a. If \(y=ux\), then \(y'=u'x+u\) and \(y''=u''x+2u'\), so

    \[\begin{aligned} x^2y''+xy'-y&=x^2(u''x+2u')+x(u'x+u)-ux\\ &=x^3u''+3x^2u'.\end{aligned}\]

    Therefore \(y=ux\) is a solution of Equation \ref{eq:5.6.12} if and only if

    \[x^3u''+3x^2u'=x^2+1, \nonumber\]

    which is a first order equation in \(u'\). We rewrite it as

    \[\label{eq:5.6.13} u''+{3\over x}u'={1\over x}+{1\over x^3}.\]

    To focus on how we apply variation of parameters to this equation, we temporarily write \(z=u'\), so that Equation \ref{eq:5.6.13} becomes

    \[\label{eq:5.6.14} z'+{3\over x}z={1\over x}+{1\over x^3}.\]

    We leave it to you to show by separation of variables that \(z_1=1/x^3\) is a solution of the complementary equation

    \[z'+{3\over x}z=0 \nonumber\]

    for Equation \ref{eq:5.6.14}. By variation of parameters, every solution of Equation \ref{eq:5.6.14} is of the form

    \[z={v\over x^3} \quad \text{where} \quad {v'\over x^3}={1\over x}+{1\over x^3}, \quad \text{so} \quad v'=x^2+1 \quad \text{and} \quad v={x^3\over 3}+x+C_1. \nonumber\]

    Since \(u'=z=v/x^3\), \(u\) is a solution of Equation \ref{eq:5.6.14} if and only if

    \[u'={v\over x^3}={1\over3}+{1\over x^2}+{C_1\over x^3}.\nonumber\]

    Integrating this yields

    \[u={x\over 3}-{1\over x}-{C_1\over2x^2}+C_2.\nonumber\]

    Therefore the general solution of Equation \ref{eq:5.6.12} is

    \[\label{eq:5.6.15} y=ux={x^2\over 3}-1-{C_1\over2x}+C_2x.\]

    Reasoning as in the solution of Example \(\PageIndex{1a}\), we conclude that \(y_1=x\) and \(y_2=1/x\) form a fundamental set of solutions for Equation \ref{eq:5.6.11}.

    As we explained above, we rename the constants in Equation \ref{eq:5.6.15} and rewrite it as

    \[\label{eq:5.6.16} y={x^2\over3}-1+c_1x+{c_2\over x}.\]

    b. Differentiating Equation \ref{eq:5.6.16} yields

    \[\label{eq:5.6.17} y'={2x\over 3}+c_1-{c_2\over x^2}.\]

    Setting \(x=1\) in Equation \ref{eq:5.6.16} and Equation \ref{eq:5.6.17} and imposing the initial conditions \(y(1)=2\) and \(y'(1)=-3\) yields

    \[\begin{aligned} c_1+c_2&= \phantom{-}{8\over 3} \\ c_1-c_2&= -{11\over 3}.\end{aligned}\]

    Solving these equations yields \(c_1=-1/2\), \(c_2=19/6\). Therefore the solution of Equation \ref{eq:5.6.12} is

    \[y={x^2\over 3}-1-{x\over 2}+{19\over 6x}.\nonumber\]

    Using reduction of order to find the general solution of a homogeneous linear second order equation leads to a homogeneous linear first order equation in \(u'\) that can be solved by separation of variables. The next example illustrates this.

    Example 5.6.3

    Find the general solution and a fundamental set of solutions of

    \[\label{eq:5.6.18} x^2y''-3xy'+3y=0,\]

    given that \(y_1=x\) is a solution.

    Solution

    If \(y=ux\) then \(y'=u'x+u\) and \(y''=u''x+2u'\), so

    \[\begin{aligned} x^2y''-3xy'+3y&=x^2(u''x+2u')-3x(u'x+u)+3ux\\ &=x^3u''-x^2u'.\end{aligned}\]

    Therefore \(y=ux\) is a solution of Equation \ref{eq:5.6.18} if and only if

    \[x^3u''-x^2u'=0.\nonumber\]

    Separating the variables \(u'\) and \(x\) yields

    \[{u''\over u'}={1\over x},\nonumber\]

    so

    \[\ln|u'|=\ln|x|+k,\quad \text{or equivalently} \quad u'=C_1x.\nonumber\]

    Therefore

    \[u={C_1\over2}x^2+C_2,\nonumber\]

    so the general solution of Equation \ref{eq:5.6.18} is

    \[y=ux={C_1\over2}x^3+C_2x,\nonumber\]

    which we rewrite as

    \[y=c_1x+c_2x^3.\nonumber\]

    Therefore \(\{x,x^3\}\) is a fundamental set of solutions of Equation \ref{eq:5.6.18}.