Question

In Problems 49-52 solve the given differential equation by using a CAS to find the (approximate) roots of the auxiliary equation.$\mathbf{50}\mathbf{.}{\mathit{x}}^{\mathbf{3}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}{\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{5}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{-}\mathbf{9}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The solution is

$y=c_1{x}^{1.408}+c_2{x}^{-1.204}cos(2.22lnx)+c_3{x}^{-1.204}sin(2.22lnx)$

Step-by-step solution

given information

$x^{3}y\text{'}\text{'}\text{'}+4x^2{y}^{\text{'}\text{'}}+5x{y}^{\text{'}}-9y=0$

to find the general solution

We have the Cauchy Euler equation

$x^{3}y\text{'}\text{'}\text{'}+4x^{2}y\text{'}\text{'}5xy\text{'}-9y=0$

and we have to obtain its general solution as the following technique:

Assume that $y=x^m$and then differentiate with respect to x, then we have

$$\begin{gathered} y^{\text{'}}=m{x}^{(m-1)}.....\left(2\right) \\ y\text{'}\text{'}=m(m-1)x^{(m-2)}.....\left(3\right) \\ y\text{'}\text{'}\text{'}=m(m-1)(m-2)x^{(m-3)}.....\left(4\right) \end{gathered}$$

Then, substitute with $y=x^m$and equations (2),(3) and (4) into equation (1), then we have

$$\begin{gathered} x^3\left(m(m-1)(m-2)x^{(m-3)}\right)-4x^2\left(m(m-1)x^{(m-2)}\right)+5x\left(m{x}^{(m-1)}\right)-9x^m=0 \\ m(m-1)(m-2)x^3\times x^{(m-3)}-4m(m-1)x^2\times x^{(m-2)}+5mx\times x^{(m-1)}-9x^m=0 \\ m(m-1)(m-2)x^m+4m(m-1)x^m+5m{x}^m-9x^m=0 \end{gathered}$$

$$\begin{gathered} \left[m\right(m-1\left)\right(m-2)+4m(m-1)+5m-9]x^m=0 \\ \left[m\left(m^2-3m+2\right)+4m^2-4m+5m-9\right]x^m=0 \\ \left(m^3-3m^2+2m-4m^2-4m+5m-9\right)x^m=0 \end{gathered}$$

$\left(m^3+m^2+3m-9\right)x^m=0$

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

$m^3+m^2+3m-9=0$

find general solution

This auxiliary equation cannot be solved easily, then using CAS we can obtain the roots for this equation as

$m_1=1.408,m_2=-1.204+2.22iand{m}_3=-1.204+2.22i$

Then we can obtain the general solution of this Cauchy Euler equation as

$\begin{array}{rcl}y& =& c_1{x}^{1.408}+c_2{\left(e^{lnx}\right)}^{(-1.204+2.22i)}+c_3{\left(e^{lnx}\right)}^{(-1.204-22.2i)}\\ & =& c_1{x}^{1.408}+c_2{e}^{(-1.204+2.22i)lnx}+c_3{e}^{(-1.204-2.22i)lnx}\\ & =& c_1{x}^{1.408}+c_2{e}^{(-1.204lnx)}{e}^{(2.22i)lnx}+c_3{e}^{(-1.204lnx)}{e}^{(-2.22i)lnx}\\ & =& c_1{x}^{1.408}+c_2{x}^{-1.204}{e}^{(2.22i)lnx}+c_3{x}^{-1.204}{e}^{(-2.22i)lnx}\\ & =& c_1{x}^{1.408}+c_2{x}^{-1.204}cos(2.22lnx)+c_3{x}^{-1.204}sin(2.22lnx)\end{array}$

conclusion

The solution is

$\begin{array}{l}y=c_1{x}^{1.408}+c_2{x}^{-1.204}cos(2.22lnx)+c_3{x}^{-1.204}sin(2.22lnx)\end{array}$