Question

In Problems 15 and 16 find a homogeneous second-order Cauchy Euler equation with real coefficients if the given numbers are roots of its auxiliary equation.
${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathit{i}$

Short answer

A homogeneous second-order Cauchy Euler equation with real coefficients is

$x^{2}y"+xy\text{'}+y=0$

Step-by-step solution

Define

In mathematics, an Euler–Cauchy equation, or Cauchy–Euler equation, or simply Euler's equation is a linear homogeneous ordinary differential equation with variable coefficients. It is sometimes referred to as an equidimensional equation.

Solve the equation.

We can obtain the Cauchy Euler equation for the given roots with real coefficients as the following technique :
First, we have to assume that

$y=x^m$

Then differentiate with respect to x , then we have

$$\begin{gathered} y\text{'}=m{x}^{\left(m-1\right)} \\ y"=m\left(m-1\right)x^{\left(m-2\right)} \end{gathered}$$

Second, since we have the roots $m_1=i$and$m_2=-i$ , then we can obtain the auxiliary equation multiplied by our assumed solution and then expand it as

$$\begin{gathered} \left(m-i\right)\left(m+i\right)x^m=0 \\ \left(m^2-i^2\right)x^m=0 \\ \left(m^2+1\right)x^m=0 \\ \left(m^2-m+m+1\right)x^m=0 \\ \left[m\left(m-1\right)+m+1\right]x^m=0 \\ \left[m\left(m-1\right)+m+1\right]x^m=0 \\ m\left(m-1\right)x^m+m{x}^m+x^m=0 \\ m\left(m-1\right)x^2.x^{\left(m-2\right)}+mx.x^{\left(m-1\right)}+x^m=0 \\ {x}^2\left(m\left(m-1\right)x^{\left(m-2\right)}\right)+x\left(m{x}^{\left(m-1\right)}\right)+\left(x^m\right)=0 \end{gathered}$$

After that, by solving the three equations (1), (2) and with equation (4), then we can have $x^{2}y"+xy\text{'}+y=0$is the required homogeneous second order Cauchy Euler equation with real coefficients.

Ahomogeneous second-order Cauchy Euler equation with real coefficients is:

$x^{2}y"+xy\text{'}+y=0$