Question

Solve the given differential equation.$\mathit{x}{\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{0}$

Short answer

The obtained solution for the given differential equation is$y=c_1+c_{2}x+c_3{x}^2+c_4{x}^{-3}$ .

Step-by-step solution

Define the general solution of 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.

Derive the given Cauchy Euler equation and find the solution

Consider the given Cauchy Euler equation,

$x{y}^{\left(4\right)}+6y\text{'}\text{'}\text{'}=0,$

Thus, the equation can be written as, $x^4{y}^{\left(4\right)}+6x^{3}y\text{'}\text{'}\text{'}=0$ (1)

Assume, the solution for this homogeneous equation is$y=x^m$

Hence, 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)

$y^{\left(4\right)}=m(m-1)(m-2)(m-3)x^{(m-4)}$ (3)

Obtain the solution by simplifying the equation

Substitute equation 2 and 3 in 1 and $y=x^m$,

$\begin{array}{rcl}{x}^4\left(m(m-1)(m-2)(m-3)x^{(m-4)}\right)+6x^3\left(m(m-1)(m-2)x^{(m-3)}\right)& =& 0\\ x^4{x}^{m-4}\left(m\right(m-1\left)\right(m-2\left)\right(m-3\left)\right)+x^3{x}^{(m-3)}\left(6m\right(m-1\left)\right(m-2\left)\right)& =& 0\\ x^m\left(m\right(m-1\left)\right(m-2\left)\right(m-3\left)\right)+x^m\left(6m\right(m-1\left)\right(m-2\left)\right)& =& 0\\ x^m\left(m\right(m-1\left)\right(m-2\left)\right(m-3)+6m(m-1\left)\right(m-2\left)\right)& =& 0\\ m(m-1)(m-2)\left[\right(m-3)+6]x^m& =& 0\\ m(m-1)(m-2)(m+3)x^m& =& 0\\ & & \end{array}$

Since$x^m$ cannot be equal to 0, we obtain:

$m(m-1)(m-2)(m+3)=0$

Thus, the roots can be,

$m_1=0m_2=1m_3=2and{m}_4=-3$

Hence, the solution for the given equation can be obtained as,

$\begin{array}{rcl}{y}_h& =& c_1{x}^0+c_2{x}^1+c_3{x}^2+c_4{x}^{-3}\\ & =& c_1+c_{2}x+c_3{x}^2+c_4{x}^{-3}\end{array}$

Therefore, the solution for the Cauchy Euler equation is$y=c_1+c_{2}x+c_3{x}^2+c_4{x}^{-3}$ .