Reduction of order

7.3. 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).

For example, to solve:

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

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 \(x^2-x\):

(7.26)#\[\begin{equation} y'' + \frac{1}{1 - x} y' + \frac{1}{x^2 - x}y = 0 \end{equation}\]

Then, calculate u:

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

Finally, integrate u to obtain \(y_2\):

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

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

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

7.3.1. Skill builder problems#

Solve the following:

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

    Solution

    Rearrange in standard form:

    (7.30)#\[\begin{equation} y'' + \frac{2}{x}y' + y = 0 \end{equation}\]

    This ODE is linear and homogeneous, so use reduction of order for \(y_2\). First compute u:

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

    Then compute \(y_2\):

    (7.32)#\[\begin{align} y_{2} &= y_{1} \int u \d{x} \\ &= \frac{\cos x}{x} \int \sec^2 x \d{x} \\ &= x^{-1}\cos x \tan x \\ &= \frac{\sin x}{x} \end{align}\]

    Hence, the general solution is:

    (7.33)#\[\begin{equation} y = c_{1}\frac{\cos x}{x} + c_{2}\frac{\sin x}{x} \end{equation}\]

  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:

    (7.34)#\[\begin{equation} y'' - \frac{2x}{1-x^2}y' + \frac{2}{1-x^2}y = 0 \end{equation}\]

    Now compute u

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

    Then \(y_2\):

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

    Use partial-fraction decomposition to evaluate the integral:

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

    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,

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

    Hence, the general solution is:

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