Question

use the procedures developed in this chapter to find the general solution of each differential equation.

$\mathbf{6}{\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{"}\mathbf{}\mathbf{+}\mathbf{5}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The solution of the given Cauchy Euler differential equation as$6x^{2}y"+5xy\text{'}-y=0asy=c_1{x}^{\frac{1}{2}}+c_2{x}^{-\frac{1}{3}}$

Step-by-step solution

Define the second order Cauchy Euler differential equation

The Cauchy-Euler differential equation is a type of linear ordinary differential equation with variable coefficients. In time-harmonic vibrations of a thin elastic rod, issues on yearly and solid discs, wave mechanics, and other domains of science and engineering, the second order Cauchy–Euler equations are applied.

${\mathit{a}}_{\mathbf{n}}{\mathit{x}}^{\mathbf{n}}{\mathit{y}}^{\left(n\right)}\mathbf{}\mathbf{+}{\mathit{a}}_{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathit{x}}^{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathit{y}}^{\left(n-1\right)}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{a}}_{\mathbf{0}}\mathit{y}\mathbf{=}\mathbf{0}$

Step 2: Obtain the solution of the given Cauchy Euler differential equation 

Here$6x^{2}y"+5x{y}^{\text{'}}-y=0$

Which is$$\begin{gathered} y"+\frac{5}{6}{x}^{-1}y\text{'}-\frac{1}{6}{x}^{-2}y=0 \\ y"+p\left(x\right)y\text{'}+q\left(x\right)y=0, \end{gathered}$$

Now obtain the solution by assuming $y=x^m$

$$\begin{gathered} y\text{'}=m{x}^{\left(m-1\right)} \\ y"=m\left(m-1\right)x^{\left(m-2\right)} \\ 6x^2\left(m\left(m-1\right)x^{\left(m-2\right)}\right)+5x\left(m{x}^{\left(m-1\right)}\right)-x^m=06m\left(m-1\right)x^2{x}^{\left(m-2\right)}+5mx{x}^{\left(m-1\right)}-x^m \\ =06m\left(m-1\right)x^m+5m{x}^m-x^m \\ =0\left(6m^2-6m+5m-1\right)x^m \\ =0\left(6m^2-m-1\right)x^m \\ =0 \end{gathered}$$

Since $x^m$ cannot be 0

then we have the auxiliary equation as

$$\begin{gathered} 6m^2-m-1=0 \\ \left(2m-1\right)\left(3m+1\right)=0 \end{gathered}$$

Roots of the equation$$\begin{gathered} m_1=\frac{1}{2} \\ {m}_2=-\frac{1}{2} \end{gathered}$$

Then we can obtain the solution of the given Cauchy Euler differential equation as

$$\begin{gathered} 6x^{2}y"+5xy\text{'}-y=0as \\ y=c_1{x}^{\frac{1}{2}}+c_2{x}^{-\frac{1}{3}} \end{gathered}$$