# 5.6: Reduction of Order

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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\\[4pt] &=(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\\[4pt] &=x^3u''+3x^2u'.\end{aligned} \nonumber

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} \\[4pt] c_1-c_2&= -{11\over 3}.\end{aligned} \nonumber

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\\[4pt] &=x^3u''-x^2u'.\end{aligned} \nonumber

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}.

This page titled 5.6: Reduction of Order is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.