5.6 Reduction of Order

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

\begin{equation} P_0(x)y''+P_1(x)y'+P_2(x)y=F(x) \tag{5.6.1}\end{equation}

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

\begin{equation} P_0(x)y''+P_1(x)y'+P_2(x)y=0. \tag{5.6.2}\end{equation}

The method is called reduction of order because it reduces the task of solving (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 (5.6.1) in the form

\begin{equation} y=uy_1 \tag{5.6.3}\end{equation}

where \(u\) is to be determined so that \(y\) satisfies (5.6.1). Substituting (5.6.3) and

\[\begin{array}{rcl} y'&=& u'y_1+uy_1' \\ y''&=& u''y_1+2u'y_1'+uy_1'' \end{array}\]

into (5.6.1) yields

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

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

\begin{equation} (P_0y_1)u''+(2P_0y_1'+P_1y_1)u'+(P_0y_1''+P_1y_1'+P_2y_1) u=F. \tag{5.6.4}\end{equation}

However, the coefficient of \(u\) is zero, since \(y_1\) satisfies (5.6.2). Therefore (5.6.4) reduces to

\begin{equation} Q_0(x)u''+Q_1(x)u'=F, \tag{5.6.5}\end{equation}

with

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

(It isn’t worthwhile to memorize the formulas for \(Q_0\) and \(Q_1\)!) Since (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 (5.6.3).

Example 5.6.1

  1. Find the general solution of

    \begin{equation} xy''-(2x+1)y'+(x+1)y=x^2, \tag{5.6.6}\end{equation}

    given that \(y_1=e^x\) is a solution of the complementary equation

    \begin{equation} xy''-(2x+1)y'+(x+1)y=0. \tag{5.6.7}\end{equation}

  2. As a byproduct of (a), find a fundamental set of solutions of (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{eqnarray*} 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{eqnarray*}

Therefore \(y=ue^x\) is a solution of (5.6.6) if and only if

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

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

\begin{equation} u''-{u'\over x}=xe^{-x}. \tag{5.6.8}\end{equation}

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

\begin{equation} z'-{z\over x}=xe^{-x}. \tag{5.6.9}\end{equation}

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

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

for (5.6.9). By applying variation of parameters as in Section 1.2, we can now see that every solution of (5.6.9) is of the form

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

Since \(u'=z=vx\), \(u\) is a solution of (5.6.8) if and only if

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

Integrating this yields

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

Therefore the general solution of (5.6.6) is

\begin{equation} y=ue^x=x+1+{C_1\over2}x^2e^x+C_2e^x. \tag{5.6.10}\end{equation}

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

Although (5.6.10) is a correct form for the general solution of (5.6.6), it’s silly to leave the arbitrary coefficient of \(x^2e^x\) as \(C_1/2\) where \(C_1\) is an arbitrary constant. Moreover, it’s 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 (5.6.10) as

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

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

    \[ x^2y''+xy'-y=x^2+1, \]

    given that \(y_1=x\) is a solution of the complementary equation

    \begin{equation} x^2y''+xy'-y=0. \tag{5.6.11}\end{equation}

    As a byproduct of this result, find a fundamental set of solutions of (5.6.11).

  2. Solve the initial value problem

    \begin{equation} x^2y''+xy'-y=x^2+1, \quad y(1)=2,\; y'(1)=-3. \tag{5.6.12}\end{equation}

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

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

Therefore \(y=ux\) is a solution of (5.6.12) if and only if

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

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

\begin{equation} u''+{3\over x}u'={1\over x}+{1\over x^3}. \tag{5.6.13}\end{equation}

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

\begin{equation} z'+{3\over x}z={1\over x}+{1\over x^3}. \tag{5.6.14}\end{equation}

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

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

for (5.6.14). By variation of parameters, every solution of (5.6.14) is of the form

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

Since \(u'=z=v/x^3\), \(u\) is a solution of (5.6.14) if and only if

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

Integrating this yields

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

Therefore the general solution of (5.6.12) is

\begin{equation} y=ux={x^2\over 3}-1-{C_1\over2x}+C_2x. \tag{5.6.15}\end{equation}

Reasoning as in the solution of Example 5.6.1(a), we conclude that \(y_1=x\) and \(y_2=1/x\) form a fundamental set of solutions for (5.6.11).

As we explained above, we rename the constants in (5.6.15) and rewrite it as

\begin{equation} y={x^2\over3}-1+c_1x+{c_2\over x}. \tag{5.6.16}\end{equation}

Solution (b) Differentiating (5.6.16) yields

\begin{equation} y'={2x\over 3}+c_1-{c_2\over x^2}. \tag{5.6.17}\end{equation}

Setting \(x=1\) in (5.6.16) and (5.6.17) and imposing the initial conditions \(y(1)=2\) and \(y'(1)=-3\) yields

\begin{eqnarray*} c_1+c_2&=& \phantom{-}{8\over 3} \\ c_1-c_2&=& -{11\over 3}. \end{eqnarray*}

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

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

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

\begin{equation} x^2y''-3xy'+3y=0, \tag{5.6.18}\end{equation}

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

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

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

Therefore \(y=ux\) is a solution of (5.6.18) if and only if

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

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

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

so

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

Therefore

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

so the general solution of (5.6.18) is

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

which we rewrite as

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

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

5.6 Exercises

In Exercises 1–17 find the general solution, given that \(y_1\) satisfies the complementary equation. As a byproduct, find a fundamental set of solutions of the complementary equation.

  1. \((2x+1)y''-2y'-(2x+3)y=(2x+1)^2; \quad y_1=e^{-x}\)

    Show answer

    \(y=1-2x+c_1e^{-x}+c_2xe^x\);  \(\{e^{-x},xe^x\}\)

  2. \(x^2y''+xy'-y=\dst{4\over x^2}; \quad y_1=x\)

    Show answer

    \(y=\dst{{4\over 3x^2}+c_1x+{c_2\over x}}\);  \(\{x,1/x\}\)

  3. \(x^2y''-xy'+y=x; \quad y_1=x\)

    Show answer

    \(y=\dst{x(\ln |x|)^2\over2}+c_1x+c_2x\ln |x|\);  \(\{x,x\ln|x|\}\)

  4. \(y''-3y'+2y=\dst{1\over1+e^{-x}}; \quad y_1=e^{2x}\)

    Show answer

    \(y=(e^{2x}+e^x) \ln(1+e^{-x})+c_1e^{2x}+c_2e^x\);  \(\{e^{2x},e^x\}\)

  5. \(y''-2y'+y=7x^{3/2}e^x; \quad y_1=e^x\)

    Show answer

    \(y=e^x\dst{\left({4\over5}x^{7/2}+c_1+c_2x\right)}\);  \(\{e^x,xe^x\}\)

  6. \(4x^2y''+(4x-8x^2)y'+(4x^2-4x-1)y=4x^{1/2}e^x(1+4x); \quad y_1=x^{1/2}e^x\)

    Show answer

    \(y=e^x(2x^{3/2}+x^{1/2}\ln x+c_1x^{1/2}+c_2x^{-1/2})\);  \(\{x^{1/2}e^x,x^{-1/2}e^{-x}\}\)

  7. \(y''-2y'+2y=e^x\sec x; \quad y_1=e^x\cos x\)

    Show answer

    \(y=e^x(x\sin x+\cos x\ln|\cos x|+c_1\cos x+c_2\sin x)\);  \(\{e^x\cos x,e^x\sin x\}\)

  8. \(y''+4xy'+(4x^2+2)y=8e^{-x(x+2)}; \quad y_1=e^{-x^2}\)

    Show answer

    \(y=e^{-x^2}(2e^{-2x}+c_1+c_2x)\);  \(\{e^{-x^2},xe^{-x^2}\}\)

  9. \(x^2y''+xy'-4y=-6x-4; \quad y_1=x^2\)

    Show answer

    \(y=2x+1+c_1x^2+\dst{c_2\over x^2}\);  \(\{x^2,1/x^2\}\)

  10. \(x^2y''+2x(x-1)y'+(x^2-2x+2)y=x^3e^{2x}; \quad y_1=xe^{-x}\)

    Show answer

    \(y=\dst{xe^{2x}\over9}+xe^{-x}(c_1+c_2x)\);  \(\{xe^{-x},x^2e^{-x}\}\)

  11. \(x^2y''-x(2x-1)y'+(x^2-x-1)y=x^2e^x; \quad y_1=xe^x\)

    Show answer

    \(y=xe^x\dst{\left({x\over 3}+c_1+{c_2\over x^2}\right)}\);  \(\{xe^x,e^x/x\}\)

  12. \((1-2x)y''+2y'+(2x-3)y=(1-4x+4x^2)e^x; \quad y_1=e^x\)

    Show answer

    \(y=-\dst{(2x-1)^2e^x\over8}+c_1e^x+c_2xe^{-x}\);  \(\{e^x,xe^{-x}\}\)

  13. \(x^2y''-3xy'+4y=4x^4; \quad y_1=x^2\)

    Show answer

    \(y=x^4+c_1x^2+c_2x^2\ln|x|\);  \(\{x^2,x^2\ln|x|\}\)

  14. \(2xy''+(4x+1)y'+(2x+1)y=3x^{1/2}e^{-x}; \quad y_1=e^{-x}\)

    Show answer

    \(y=e^{-x}(x^{3/2}+c_1+c_2x^{1/2})\);   \(\{e^{-x},x^{1/2}e^{-x}\}\)

  15. \(xy''-(2x+1)y'+(x+1)y=-e^x; \quad y_1=e^x\)

    Show answer

    \(y=e^x(x+c_1+c_2x^2)\);  \(\{e^x,x^2e^x\}\)

  16. \(4x^2y''-4x(x+1)y'+(2x+3)y=4x^{5/2}e^{2x}; \quad y_1=x^{1/2}\)

    Show answer

    \(y=x^{1/2}\dst{\left({e^{2x}\over2}+c_1+c_2e^x\right)}\);  \(\{x^{1/2},x^{1/2}e^x\}\)

  17. \(x^2y''-5xy'+8y=4x^2; \quad y_1=x^2\)

    Show answer

    \(y=-2x^2\ln x+c_1x^2+c_2x^4\);  \(\{x^2,x^4\}\)

In Exercises 18–30 find a fundamental set of solutions, given that \(y_1\) is a solution.

  1. \(xy''+(2-2x)y'+(x-2)y=0; \quad y_1=e^x\)

    Show answer

    \(\{e^x,e^x/x\}\)

  2. \(x^2y''-4xy'+6y=0; \quad y_1=x^2\)

    Show answer

    \(\{x^2,x^3\}\)

  3. \(x^2(\ln |x|)^2y''-(2x \ln |x|)y'+(2+\ln |x|)y=0; \quad y_1=\ln |x|\)

    Show answer

    \(\{\ln|x|,x \ln|x|\}\)

  4. \(4xy''+2y'+y=0; \quad y_1=\sin \sqrt{x}\)

    Show answer

    \(\{\sin\sqrt x,\cos \sqrt{x}\}\)

  5. \(xy''-(2x+2)y'+(x+2)y=0; \quad y_1=e^x\)

    Show answer

    \(\{e^x,x^3e^x\}\)

  6. \(x^2y''-(2a-1)xy'+a^2y=0; \quad y_1=x^a\)

    Show answer

    \(\{x^a,x^a\ln x\}\)

  7. \(x^2y''-2xy'+(x^2+2)y=0; \quad y_1=x \sin x\)

    Show answer

    \(\{x\sin x,x\cos x\}\)

  8. \(xy''-(4x+1)y'+(4x+2)y=0; \quad y_1=e^{2x}\)

    Show answer

    \(\{e^{2x},x^2e^{2x}\}\)

  9. \(4x^2(\sin x)y''-4x(x\cos x+\sin x)y'+(2x\cos x+3\sin x)y=0; \quad y_1=x^{1/2}\)

    Show answer

    \(\{x^{1/2},x^{1/2}\cos x\}\)

  10. \(4x^2y''-4xy'+(3-16x^2)y=0; \quad y_1=x^{1/2}e^{2x}\)

    Show answer

    \(\{x^{1/2}e^{2x},x^{1/2}e^{-2x}\}\)

  11. \((2x+1)xy''-2(2x^2-1)y'-4(x+1)y=0; \quad y_1=1/x\)

    Show answer

    \(\{1/x,e^{2x}\}\)

  12. \((x^2-2x)y''+(2-x^2)y'+(2x-2)y=0; \quad y_1=e^x\)

    Show answer

    \(\{e^x,x^2\}\)

  13. \(xy''-(4x+1)y'+(4x+2)y=0; \quad y_1=e^{2x}\)

    Show answer

    \(\{e^{2x},x^2e^{2x}\}\)

In Exercises 31–33 solve the initial value problem, given that \(y_1\) satisfies the complementary equation.

  1. \(x^2y''-3xy'+4y=4x^4,\quad y(-1)=7,\quad y'(-1)=-8; \quad y_1=x^2\)

    Show answer

    \(y=x^4+6x^2-8x^2\ln|x|\)

  2. \((3x-1)y''-(3x+2)y'-(6x-8)y=0, \quad y(0)=2,\; y'(0)=3; \quad y_1=e^{2x}\)

    Show answer

    \(y=2e^{2x}-xe^{-x}\)

  3. \((x+1)^2y''-2(x+1)y'-(x^2+2x-1)y=(x+1)^3e^x, \quad y(0)=1,\quad y'(0)=~-1\);

    \(y_1=(x+1)e^x\)

    Show answer

    \(y=\dst{(x+1)\over4}\left[-e^x(3-2x)+7e^{-x}\right]\)

In Exercises 34 and 35 solve the initial value problem and graph the solution, given that \(y_1\) satisfies the complementary equation.

  1. C/G \(x^2y''+2xy'-2y=x^2, \quad y(1)=\dst{5\over4},\; y'(1)=\dst{3\over2}; \quad y_1=x\)

    Show answer

    \(y=\dst{x^2\over4}+x\)

  2. C/G \((x^2-4)y''+4xy'+2y=x+2, \quad y(0)=-\dst{1\over3},\quad y'(0)=-1; \quad y_1=\dst{1\over x-2}\)

    Show answer

    \(y=\dst{{(x+2)^2\over6(x-2)}+{2x\over x^2-4}}\)

  3. Suppose \(p_1\) and \(p_2\) are continuous on \((a,b)\). Let \(y_1\) be a solution of

    \[ y''+p_1(x)y'+p_2(x)y=0 \tag*{\rm (A)} \]

    that has no zeros on \((a,b)\), and let \(x_0\) be in \((a,b)\). Use reduction of order to show that \(y_1\) and

    \[ y_2(x)=y_1(x)\int^x_{x_0}{1\over y^2_1(t)} \exp \left(-\int^t_{x_0}p_1(s)\, ds\right)\,dt \]

    form a fundamental set of solutions of (A) on \((a,b)\). (Note: This exercise is related to Exercise 9.)

  4. The nonlinear first order equation

    \[ y'+y^2+p(x)y+q(x)=0 \tag*{\rm (A)} \]

    is a Riccati equation. (See Exercise 2.4. 55.) Assume that \(p\) and \(q\) are continuous.

    1. Show that \(y\) is a solution of (A) if and only if \(y={z'/z}\), where

      \[ z''+p(x)z'+q(x)z=0. \tag*{\rm (B)} \]

    2. Show that the general solution of (A) is

      \[ y={c_1z'_1+c_2z'_2\over c_1z_1+c_2z_2}, \tag*{\rm (C)} \]

      where \(\{z_1,z_2\}\) is a fundamental set of solutions of (B) and \(c_1\) and \(c_2\) are arbitrary constants.

    3. Does the formula (C) imply that the first order equation (A) has a two–parameter family of solutions? Explain your answer.

  5. Use a method suggested by Exercise 37 to find all solutions. of the equation.

    (a) \(y'+y^2+k^2=0\)(b) \(y'+y^2-3y+2=0\)
    (c) \(y'+y^2+5y-6=0\)(d) \(y'+y^2+8y+7=0\)
    (e) \(y'+y^2+14y+50=0\)(f) \(6y'+6y^2-y-1=0\)

    (g) \(36y'+36y^2-12y+1=0\)

    Show answer

    (a) \(y=\dst{-kc_1\sin kx+kc_2\cos kx\over c_1 \cos kx+c_2\sin kx}\)   (b) \(y=\dst{c_1+2c_2e^x\over c_1+c_2e^x}\)
    [2]

        (c) \(y=\dst{-6c_1+c_2e^{7x}\over c_1+c_2e^{7x}}\)   (d) \(y=-\dst{7c_1+c_2e^{6x}\over c_1+c_2e^{6x}}\)

        (e) \(y=-\dst{(7c_1-c_2)\cos x+(c_1+7c_2)\sin x\over c_1\cos x+c_2\sin x}\)

        (f) \(y=\dst{-2c_1+3c_2e^{5x/6}\over 6(c_1+c_2e^{5x/6})}\)   (g) \(y=\dst{c_1+c_2(x+6)\over6(c_1+c_2x)}\)

  6. Use a method suggested by Exercise 37 and reduction of order to find all solutions of the equation, given that \(y_1\) is a solution.

    1. \(x^2(y'+y^2)-x(x+2)y+x+2=0; \quad y_1=1/x\)

    2. \(y'+y^2+4xy+4x^2+2=0; \quad y_1=-2x\)

    3. \((2x+1)(y'+y^2)-2y-(2x+3)=0; \quad y_1=-1\)

    4. \((3x-1)(y'+y^2)-(3x+2)y-6x+8=0; \quad y_1=2\)

    5. \(\dst{x^2(y'+y^2)+xy+x^2- {1\over 4}=0; \quad y_1=-\tan x -{1\over 2x}}\)

    6. \(\dst{x^2(y'+y^2)-7xy+7=0; \quad y_1=1/x}\)

    Show answer

    (a) \(y=\dst{{c_1+c_2e^x(1+x)\over x(c_1+c_2e^x)}}\)  (b) \(y=\dst{{-2c_1x+c_2(1-2x^2)\over c_1+c_2x}}\)
    [2]

        (c) \(y=\dst{{-c_1+c_2e^{2x}(x+1)\over c_1+c_2xe^{2x}}}\)  (d) \(y=\dst{{2c_1+c_2e^{-3x}(1-x)\over c_1+c_2xe^{-3x}}}\)

        (e) \(y=\dst{{(2c_2x-c_1)\cos x-(2c_1x+c_2)\sin x \over 2x(c_1\cos x+c_2\sin x)}}\)   (f) \(y=\dst{c_1+7c_2x^6\over x(c_1+c_2x^6)}\)

  7. The nonlinear first order equation

    \[ y'+r(x)y^2+p(x)y+q(x)=0 \tag*{\rm (A)} \]

    is the generalized Riccati equation. (See Exercise 2.4. 55.) Assume that \(p\) and \(q\) are continuous and \(r\) is differentiable.

    1. Show that \(y\) is a solution of (A) if and only if \(y={z'/rz}\), where

      \[ z''+\left[p(x)-{r'(x)\over r(x)}\right] z'+r(x)q(x)z=0. \tag*{\rm (B)} \]

    2. Show that the general solution of (A) is

      \[ y={c_1z'_1+c_2z'_2\over r(c_1z_1+c_2z_2)}, \]

      where \(\{z_1,z_2\}\) is a fundamental set of solutions of (B) and \(c_1\) and \(c_2\) are arbitrary constants.