Question

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

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

Short answer

The solution of the given Cauchy Euler differential equation

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

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.

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

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

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

Which is$y^{\text{'}\text{'}}+\frac{5}{6}{x}^{-1}{y}^{\text{'}}-\frac{1}{6}{x}^{-2}y=0$

$y^{\text{'}\text{'}}+p\left(x\right)y^{\text{'}}+q\left(x\right)y=0,$

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

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

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

Since $x^m$cannot be 0

then we have the auxiliary equation as

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

Roots of the equation

$$\begin{gathered} m_1=\frac{1}{2} \\ {m}_2=-\frac{1}{3} \end{gathered}$$

Then we can obtain the solution of the given Cauchy Euler differential equation $6x^2{y}^{\text{'}\text{'}}+5x{y}^{\text{'}}-y=0$as

$y=c_1{x}^{\frac{1}{2}}+c_2{x}^{-\frac{1}{3}}$