Question

In Problems 49-52 solve the given differential equation by using a CAS to find the (approximate) roots of the auxiliary equation.$\mathbf{51}\mathbf{.}{\mathit{x}}^{\mathbf{4}}{\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{+}\mathbf{6}{\mathit{x}}^{\mathbf{3}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{3}{\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{3}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The solution is: $y=c_1{x}^{1.414}+c_2{x}^{-1.414}+c_3{x}^{1.414}lnx+c_2{x}^{-1.414}lnx$

Step-by-step solution

given information

$x^4{y}^{\left(4\right)}+6x^3{y}^{\text{'}\text{'}\text{'}}+3x^2{y}^{\text{'}\text{'}}-3x{y}^{\text{'}}+4y=0$

To find the general solution

We have the Cauchy Euler equation:

$x^4{y}^{\left(4\right)}+6x^3{y}^{\text{'}\text{'}\text{'}}+3x^2{y}^{\text{'}\text{'}}-3x{y}^{\text{'}}+4y=0$

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

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

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

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

$$\begin{gathered} x^4\left(m(m-1)(m-2)(m-3)x^{(m-4)}\right)+6x^3\left(m(m-1)(m-2)x^{(m-3)}\right)+3x^2\left(m(m-1)x^{(m-2)}\right)-3x\left(m{x}^{(m-1)}\right)+4x^m=0 \\ (m-1)(m-2)(m-3)x^4\times x^{(m-4)}+6m(m-1)(m-2)x^3\times x^{(m-3)}+3m(m-1)x^2\times x^{(m-2)}-3mx\times x^{(m-1)}+4x^m=0 \\ m(m-1)(m-2)(m-3)x^m+6m(m-1)(m-2)x^m+3m(m-1)x^m-3m{x}^m+4x^m=0 \end{gathered}$$

Now,

$$\begin{gathered} \left[m\right(m-1\left)\right(m-2\left)\right(m-3)+6m(m-1\left)\right(m-2)+3m(m-1)-3m+4]x^m=0 \\ \left[\left(m^2-m\right)\left(m^2-5m+6\right)+6m\left(m^2-3m+2\right)+\left(3m^2-3m\right)-3m+4\right]x^m=0 \\ \left(m^4-6m^3+11m^2-6m+6m^3-18m^2+12m+3m^2-3m-3m+4\right)x^m=0 \end{gathered}$$

Now,

$\left(m^4-4m^2+4\right)x^m=0$

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

$$\begin{gathered} m^4-4m^2+4=0 \\ \left(m^2-2\right)\left(m^2-2\right)=0 \end{gathered}$$

Find the general solution

Then the roots are:

$m_1=\sqrt{2},m_2=-\sqrt{2},m_3=\sqrt{2}$

and$m_4=-\sqrt{2}$

Since we have repeated roots $m_{1,3}=\sqrt{2}$and $m_{2,4}=-\sqrt{2}$, then we have to add an additional solution for each root, then we can obtain the general solution of this Cauchy Euler equation as:

$\begin{array}{rcl}y& =& c_1{x}^{\sqrt{2}}+c_2{x}^{-\sqrt{2}}+c_3{x}^{\sqrt{2}}lnx+c_2{x}^{-\sqrt{2}}lnx\\ & =& c_1{x}^{1.414}+c_2{x}^{-1.414}+c_3{x}^{1.414}lnx+c_2{x}^{-1.414}lnx\\ & & \end{array}$

Conclusion

The solution is:

$y=c_1{x}^{1.414}+c_2{x}^{-1.414}+c_3{x}^{1.414}lnx+c_2{x}^{-1.414}lnx$