# Reduction of order

It can be hard to find $y_1$ and $y_2$ in general, but some techniques can help.
We will learn how to get $y_2$ if a $y_1$ is found (e.g, by guessing).

```{topic} Reduction of Order
If a solution $y_1$ of $y'' + p(x) y' + q(x) y = 0$ is known, then

\begin{align}
y_2 &= y_1\int u\d{x} \\
u &= \frac{1}{y_1^2}\ e^{-\int p\d{x}}
\end{align}

is another solution.
```

For example, to solve:

\begin{equation}
(x^2 - x)y'' - xy' + y = 0
\end{equation}

We can guess that $y_1 = x$ is a solution because $y_1' = 1$ and $y_1'' = 0$.
We can now use reduction of order to find $y_2$. First, rewrite the ODE in
standard form by dividing through by $x^2-x$:

\begin{equation}
y'' + \frac{1}{1 - x} y' + \frac{1}{x^2 - x}y = 0
\end{equation}

Then, calculate *u*:

\begin{align}
u &= \frac{1}{x^2} \exp\left(-\int \frac{1}{1-x} \d{x} \right) \\
  &= \frac{1}{x^2}\ e^{\ln(1-x)} \\
  &= \frac{1-x}{x^2}
\end{align}

Finally, integrate *u* to obtain $y_2$:

\begin{align}
y_2 &= \frac{1}{x^2} \int \left( \frac{1}{x^2} - \frac{1}{x} \right)\d{x} \\
&= x \left(-\frac{1}{x}-\ln{x}\right)
\end{align}

Absorbing the minus sign into the unknown coefficient, the general solution is:

\begin{equation}
y = c_1 x + c_2 (1 + \ln x)
\end{equation}

## Skill builder problems

Solve the following:

1. $xy'' + 2y' + xy = 0$, given $y_1 = \cos x / x$

   ```{solution}
   Rearrange in standard form:

   \begin{equation}
   y'' + \frac{2}{x}y' + y = 0
   \end{equation}

   This ODE is linear and homogeneous, so use reduction of order for $y_2$.
   First compute *u*:

   \begin{align}
   u &= \frac{1}{y_1^2}e^{-\int p \d{x}} \\
     &= \frac{x^2}{\cos^2 x} \exp\left(-\int \frac{2}{x} \d{x}\right) \\
     &= \frac{x^2}{\cos^2 x} \exp\left(-2\ln(x)\right) \\
     &= \frac{x^2}{\cos^2 x}x^{-2} \\
     &= \sec^2 x
   \end{align}

   Then compute $y_2$:

   \begin{align}
   y_{2} &= y_{1} \int u \d{x} \\
         &= \frac{\cos x}{x} \int \sec^2 x \d{x} \\
         &= x^{-1}\cos x \tan x \\
         &= \frac{\sin x}{x}
   \end{align}

   Hence, the general solution is:

   \begin{equation}
   y = c_{1}\frac{\cos x}{x} + c_{2}\frac{\sin x}{x}
   \end{equation}
   ```

2. $(1-x^2)y'' - 2xy' + 2y = 0$

   ```{solution}
   By inspection $y_{1} = x$ is a solution because $y_{1}' = 1$ and $y_{1}'' = 0$.
   So, find $y_{2}$ using reduction of order. The standard form of the ODE is:

   \begin{equation}
   y'' - \frac{2x}{1-x^2}y' + \frac{2}{1-x^2}y = 0
   \end{equation}

   Now compute *u*

   \begin{align}
   u &= \frac{1}{y_{2}^2}e^{-\int p \d{x}} \\
     &= \frac{1}{x^2} \exp\left(\int \frac{2x}{1-x^2} \d{x}\right) \\
     &= \frac{1}{x^2} \exp\left[-\ln(1 - x^2)\right] \\
     &= \frac{1}{x^2(1 - x^2)}
   \end{align}

   Then $y_2$:

   \begin{align}
   y_{2} &= y_{1} \int u \d{x} \\
         &= x \int \frac{1}{x^2(1 - x^2)} \d{x}
   \end{align}

   Use partial-fraction decomposition to evaluate the integral:

   \begin{equation}
   \frac{1}{x^2(1 - x)(1 + x)} = \frac{A}{x} + \frac{B}{x^2} +
   \frac{C}{1 + x} + \frac{D}{1 - x}
   \end{equation}

   Use the coverup method to get $B = 1$, $C = 1/2$, and $D = 1/2$. Cross
   multiplying and collecting terms shows that $A = 0$. So,

   \begin{align}
   y_2 &= x \int \left[ \frac{1}{x^2} + \frac{1}{2(1 + x)}
         + \frac{1}{2(1 - x)} \right] \d{x} \\
       &= x \left[ -\frac{1}{x} + \frac{1}{2}\ln(1 + x)
   - \frac{1}{2}\ln(1 - x) \right] \\
       &= -1 - \frac{x}{2} \ln \left( \frac{1 - x}{1 + x} \right)
   \end{align}

   Hence, the general solution is:

   \begin{equation}
   y = c_{1}x + c_{2} \left[ 1 + \frac{x}{2}
   \ln \left( \frac{1 - x}{1 + x} \right) \right]
   \end{equation}
   ```