Question

In Problems 49-52 solve the given differential equation by using a CAS to find the (approximate) roots of the auxiliary equation.$\mathbf{2}{\mathit{x}}^{\mathbf{3}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{10}\mathbf{.}\mathbf{98}{\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{8}\mathbf{.}\mathbf{5}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{3}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The solution is $y=c_1{x}^{6.732}+c_2{x}^{1.811}+c_3{x}^{-0.0533}$

Step-by-step solution

given information

$2x^3{y}^{\text{'}\text{'}}-10.98x^2{y}^{\text{'}\text{'}}+8.5x{y}^{\text{'}}+1.3y=0$

to find the general solution

We have the Cauchy Euler equation

$2x^3{y}^{\text{'}\text{'}\text{'}}-10.98x^2{y}^{\text{'}\text{'}}+\text{8.5}x{y}^{\text{'}}+\text{1.3}y=0$

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

$$\begin{gathered} 2x^3\left(m(m-1)(m-2)x^{(m-3)}\right)-10.98x^2\left(m(m-1)x^{(m-2)}\right)+8.5x\left(m{x}^{(m-1)}\right)+1.3x^m=0 \\ 2m(m-1)(m-2)x^3\times x^{(m-3)}-10.98m(m-1)x^2\times x^{(m-2)}+8.5mx\times x^{(m-1)}+1.3x^m=0 \\ 2m(m-1)(m-2)x^m-10.98m(m-1)x^m+8.5m{x}^m+1.3x^m=0 \end{gathered}$$

Now

$$\begin{gathered} \left[2m\right(m-1\left)\right(m-2)-10.98m(m-1)+8.5m+1.3]x^m=0 \\ \left[2m\left(m^2-3m+2\right)-10.98m^2+10.98m+8.5m+1.3\right]x^m=0 \\ \left(2m^3-6m^2+4m-10.98m^2+10.98m+8.5m+1.3\right)x^m=0 \end{gathered}$$

$\left(2m^3-16.98m^2+23.48m+1.3\right)x^m=0$

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

$2m^3-16.98m^2-0.48m+1.3=0$

find general solution

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

$m_1=6.732,m_2=1.811$and $m_3=-0.0533$

Then we can obtain the general solution of this Cauchy Euler equation as${\text{y =c}}_1{\text{x}}^{\text{6.732}}{\text{+c}}_2{\text{x}}^{\text{1.811}}{\text{+c}}_3{\text{x}}^{\text{- 0.0533}}$

conclusion

The solution is ${\text{y =c}}_1{\text{x}}^{\text{6.732}}{\text{+c}}_2{\text{x}}^{\text{1.811}}{\text{+c}}_3{\text{x}}^{\text{- 0.0533}}$