Question

Solve the given differential equation.${\mathit{x}}^{\mathbf{3}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The required solution for the given differential equation is$y=c_1{x}^3+c_2\text{cos}\left(\sqrt{2}\text{ln}x\right)+c_3\text{sin}\left(\sqrt{2}\text{ln}x\right)$ .

Step-by-step solution

Define Cauchy Euler equation:

Cauchy Euler equation is a differential equation that can be in the form,$a_n{x}^n\frac{d^{n}y}{d{x}^n}+a_{n-1}{x}^{n-1}\frac{d^{n-1}y}{d{x}^{n-1}}+\cdots +a_{1}x\frac{dy}{dx}+a_{0}y=g\left(x\right)$

, here $a_n,a_{n-1},\dots ,a_0$are constants.

Determine the solution by differentiating the equation:

Consider the given Cauchy Euler equation,

$x^{3}y\text{'}\text{'}\text{'}-6y=0$ (1)

Let $y=x^m$is a solution.

Differentiate$y=x^m$ with respect to x

$y\text{'}=m{x}^{(m-1)}$

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

$y\text{'}\text{'}\text{'}=m(m-1)(m-2)x^{(m-3)}$ (2)

Substitute equation (2) into (1) and replace $y=x^m$ in equation (1),

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

Hence, $x^m$cannot be equal to 0 the equation can be written as

$m^3-3m^2+2m-6=0$

Substitute the values and obtain the solution for the given Cauchy Euler equation

Substitute$m=3$ in the equation.

Thus, it will be found that m-3 is a factor.

Hence, by making long division, the second part of a equation can be obtained as,

$(m-3)\left(m^2+2\right)=0$

Therefore, the roots will be $m_1=3and{m}_{2,3}=\pm \sqrt{2}i$,

Obtain the solution of the given equation as,

$\begin{array}{rcl}{y}_h& =& c_1{x}^3+c_2{x}^{\sqrt{2}i}+c_3{x}^{-\sqrt{2}i}\\ & =& c_1{x}^3+c_2{\left(e^{\ln x}\right)}^{\sqrt{2}i}+c_3{\left(e^{\ln x}\right)}^{-\sqrt{2}i}\\ & =& c_1{x}^3+c_2\left(e^{\sqrt{2}i\ln x}\right)+c_3\left(e^{-\sqrt{2}i\ln x}\right)\\ y& =& c_1{x}^3+c_2\text{cos}\left(\sqrt{2}\text{ln}x\right)+c_3\text{sin}\left(\sqrt{2}\text{ln}x\right)\\ & & \end{array}$

Thus, the solution for the given differential equation can be obtained as $y=c_1{x}^3+c_2\text{cos}\left(\sqrt{2}\text{ln}x\right)+c_3\text{sin}\left(\sqrt{2}\text{ln}x\right)$.