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4.2 Reduction of Order: Class Notes

4.2a

Consider a 2nd-order DE

$$a_2(x)y″+a_1(x)y\prime +a_0(x)y=0(1)$$

If $y_1,y_2$ form a fundamental set of solutions, then they are linearly independent. If $c_1{y}_1+c_2{y}_2=0$ with $c_1$ and $c_2\ne 0$, we could write

$${\displaystyle y_2=-\frac{c_1}{c_2}{y}_1.}$$

Because $y_1$ and $y_2$ are linearly independent, the multiplier ${\displaystyle -\frac{c_1}{c_2}}$ cannot be a constant; it must be a function of $x$. So we look for

$$y_2=u(x)y_1,$$

where $u(x)$ is an unknown function of $x$.

In standard form, (1) becomes

$$y″+P(x)y\prime +Q(x)y=0.$$

If $y_1$ is a known solution, consider $y_2=u{y}_1$. By the product rule,

$$y_2\prime =u\prime y_1+u{y}_1\prime ,y_2″=u″y_1+2u\prime y_1\prime +u{y}_1″.$$

Substitute $y_2″$, $y_2\prime$, and $y_2=u{y}_1$ into the standard form:

$$u″y_1+2u\prime y_1\prime +u{y}_1″+P(x)\left(u\prime y_1+u{y}_1\prime \right)+Q(x)u{y}_1=0.$$

The three terms containing $u$ itself (not its derivatives) group together, and their sum is zero because $y_1$ is a solution:

$$u(y_1″+P(x)y_1\prime +Q(x)y_1)=0\text{(}{y}_1\text{ is a solution).}$$

What remains is

$$\begin{array}{cc}\hfill {\displaystyle u″y_1+2u\prime y_1\prime +Pu\prime y_1}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle u″y_1+(2y_1\prime +P{y}_1)u\prime }& {\displaystyle =0.}\hfill \end{array}$$

Let $w=u\prime$. Then $w\prime =u″$, and the equation becomes first order and linear in $w$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle w\prime y_1+(2y_1\prime +P{y}_1)w}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle w\prime +(2\frac{y_1^{\prime }}{y_1}+P)w}& {\displaystyle =0.}\hfill \end{array}}$$

Integrating factor (since ${\displaystyle \int 2\frac{y_1^{\prime }}{y_1}dx=2\ln |y_1|=\ln y_1^2}$):

$${\displaystyle \mu =e^{\int (2\frac{y_1^{\prime }}{y_1}+P)dx}=e^{\ln y_1^2}{e}^{\int Pdx}=y_1^2{e}^{\int Pdx}.}$$

So

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \frac{d}{dx}\left(y_1^2{e}^{\int Pdx}w\right)}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle y_1^2{e}^{\int Pdx}w}& {\displaystyle =c.}\hfill \end{array}}$$

We only need one $y_2$, so let $c=1$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle u\prime =w}& {\displaystyle =\frac{e^{-\int Pdx}}{y_1^2}}\hfill \\ \hfill {\displaystyle u}& {\displaystyle =\int \frac{e^{-\int Pdx}}{y_1^2}dx.}\hfill \end{array}}$$

Then $y_2=u{y}_1$, and the general solution is $y=c_1{y}_1+c_2{y}_2$.

The reduction of order formula is

$${\displaystyle y_2=y_1\int \frac{e^{-\int P(x)dx}}{y_1^2}dx}$$

4.2b

Example

The indicated function $y_1(x)$ is a solution of the given differential equation. Use reduction of order (the process or the formula) to find a second solution $y_2(x)$.

$$y″+9y=0;y_1=\sin 3x$$

Solution

Process. Let $y_2=u(x)y_1$. Here, $y_2=u\sin 3x$. By the product rule,

$$\begin{array}{cc}\hfill {\displaystyle y_2\prime }& {\displaystyle =u\prime \sin 3x+3u\cos 3x}\hfill \\ \hfill {\displaystyle y_2″}& {\displaystyle =u″\sin 3x+3u\prime \cos 3x+3u\prime \cos 3x-9u\sin 3x}\hfill \\ \hfill {\displaystyle }& {\displaystyle =u″\sin 3x+6u\prime \cos 3x-9u\sin 3x.}\hfill \end{array}$$

Substitute into $y″+9y=0$. The $u$ terms cancel:

$$\begin{array}{cc}\hfill {\displaystyle u″\sin 3x+6u\prime \cos 3x-9u\sin 3x+9u\sin 3x}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle u″\sin 3x+6u\prime \cos 3x}& {\displaystyle =0.}\hfill \end{array}$$

Let $w=u\prime$, so $w\prime =u″$. The equation is first order in $w$ (linear, and also separable):

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle w\prime \sin 3x+6w\cos 3x}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle w\prime +6\frac{\cos 3x}{\sin 3x}w}& {\displaystyle =0.}\hfill \end{array}}$$

$${\displaystyle \mu =e^{6\int \frac{\cos 3x}{\sin 3x}dx}=e^{2\ln |\sin 3x|}={\sin }^{2}3x}$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \frac{d}{dx}({\sin }^{2}3xw)}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle {\sin }^{2}3xw}& {\displaystyle =c}\hfill \\ \hfill {\displaystyle u\prime =w}& {\displaystyle =c{\csc }^{2}3x}\hfill \\ \hfill {\displaystyle u}& {\displaystyle =c\int {\csc }^{2}3xdx=-\frac{1}{3}c\cot 3x+C_1.}\hfill \end{array}}$$

Then

$${\displaystyle y_2=u{y}_1=-\frac{1}{3}c\cot 3x\sin 3x=-\frac{1}{3}c\cos 3x.}$$

Tip: We don't need the constants. The $C_1$ term would only give back a multiple of $y_1$, and the constant ${\displaystyle -\frac{1}{3}c}$ is absorbed into $c_2$ in the general solution, so take $y_2=\cos 3x$.

$$y=c_1\sin 3x+c_2\cos 3x$$

Example

$6y″+y\prime -y=0;y_1=e^{x/3}$

Solution

Formula. First put the equation in standard form (divide by 6):

$${\displaystyle y″+\frac{1}{6}y\prime -\frac{1}{6}y=0\Rightarrow P(x)=\frac{1}{6}.}$$

Here ${\displaystyle e^{-\int P(x)dx}=e^{-x/6}}$ and $y_1^2=e^{2x/3}$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle y_2}& {\displaystyle =y_1\int \frac{e^{-\int Pdx}}{y_1^2}dx=y_1\int \frac{e^{-x/6}}{e^{2x/3}}dx}\hfill \\ \hfill {\displaystyle }& {\displaystyle =y_1\int e^{-5x/6}dx}\hfill \\ \hfill {\displaystyle }& {\displaystyle =y_1(-\frac{6}{5}{e}^{-5x/6}+C).}\hfill \end{array}}$$

We don't need these constants (${\displaystyle -\frac{6}{5}}$ and $C$), so

$$y_2=e^{x/3}{e}^{-5x/6}=e^{-x/2},y=c_1{e}^{x/3}+c_2{e}^{-x/2}.$$

4.2c

Example

$x^{2}y″-3xy\prime +5y=0;y_1=x^2\cos (\ln x)$

Solution

Standard form (divide by $x^2$):

$${\displaystyle y″-\frac{3}{x}y\prime +\frac{5}{x^2}y=0\Rightarrow P(x)=-\frac{3}{x}.}$$

Then, using the log rule $3\ln x=\ln x^3$,

$${\displaystyle e^{-\int P(x)dx}=e^{\int \frac{3}{x}dx}=e^{3\ln x}=e^{\ln x^3}=x^3.}$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle y_2}& {\displaystyle =x^2\cos (\ln x)\int \frac{x^3}{x^4{\cos }^2(\ln x)}dx}\hfill \\ \hfill {\displaystyle }& {\displaystyle =x^2\cos (\ln x)\int \frac{1}{x}{\sec }^2(\ln x)dx.}\hfill \end{array}}$$

Substitute $u=\ln x$, ${\displaystyle du=\frac{1}{x}dx}$:

$$\begin{array}{cc}\hfill {\displaystyle y_2}& {\displaystyle =x^2\cos (\ln x)\tan (\ln x)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =x^2\sin (\ln x).}\hfill \end{array}$$

Example

The indicated function $y_1(x)$ is a solution of the associated homogeneous equation. Use the method of reduction of order [the process] to find a second solution $y_2(x)$ of the homogeneous equation and a particular solution $y_p(x)$ of the given nonhomogeneous equation.

$$y″-4y\prime +3y=x;y_1=e^x$$

Dr. Matt's recommendation: for nonhomogeneous DEs, use the process, not the formula.

Solution

The process starts with $y_2=u{y}_1$, but for a nonhomogeneous equation we find that $u{y}_1$ contains both pieces:

$$u{y}_1=C{y}_2+y_p.$$

Let $y_2=u{e}^x$. Then

$$y_2\prime =u\prime e^x+u{e}^x,y_2″=u″e^x+2u\prime e^x+u{e}^x.$$

Substitute into $y″-4y\prime +3y=x$. The $u{e}^x$ terms cancel $(1-4+3=0)$:

$$\begin{array}{cc}\hfill {\displaystyle u″e^x+2u\prime e^x+u{e}^x-4u\prime e^x-4u{e}^x+3u{e}^x}& {\displaystyle =x}\hfill \\ \hfill {\displaystyle u″e^x-2u\prime e^x}& {\displaystyle =x.}\hfill \end{array}$$

Let $w=u\prime$:

$$w\prime e^x-2w{e}^x=x\Rightarrow w\prime -2w=x{e}^{-x}.$$

This is linear with integrating factor $\mu =e^{-2x}$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \int \frac{d}{dx}\left(e^{-2x}w\right)dx}& {\displaystyle =\int x{e}^{-3x}dx}\hfill \\ \hfill {\displaystyle e^{-2x}w}& {\displaystyle =-\frac{1}{3}x{e}^{-3x}-\frac{1}{9}{e}^{-3x}+C}\hfill \\ \hfill {\displaystyle u\prime =w}& {\displaystyle =-\frac{1}{3}x{e}^{-x}-\frac{1}{9}{e}^{-x}+C{e}^{2x}}\hfill \end{array}}$$

Integrate again (by parts for the first term; the constant $\frac{C}{2}$ is renamed $C$):

$${\displaystyle u=\frac{1}{3}x{e}^{-x}+\frac{4}{9}{e}^{-x}+C{e}^{2x}.}$$

Multiply by $y_1=e^x$ to get $u{y}_1$:

$${\displaystyle u{y}_1=\frac{1}{3}x+\frac{4}{9}+C{e}^{3x}.}$$

The term with the arbitrary constant is the second homogeneous solution, and the rest is the particular solution.