5.6: Reduction of Order

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 $\ref{eq:5.6.18}$.