Question

Solve the given differential equation.${\mathit{x}}^{\mathbf{4}}{\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{+}\mathbf{6}{\mathit{x}}^{\mathbf{3}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{9}{\mathit{x}}^{\mathbf{3}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The solution for the given differential equation is $y=c_1\text{cos}\left(\text{ln}x\right)+c_2\text{sin}\left(\text{ln}x\right)+c_3\text{ln}x\text{cos}\left(\text{ln}x\right)+c_4\text{ln}x\text{sin}\left(\text{ln}x\right)$.

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.

Solve the given equation by simplifying the values

Consider the given equation,

$x^4{y}^{\left(4\right)}+6x^{3}y\text{'}\text{'}\text{'}+9x^{2}y\text{'}\text{'}+3xy\text{'}+y=0$ (1)

Hence, find the solution by assuming$y=x^m$ ,

Thus, differentiate the equation $y=x^m$ with respect to x

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

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

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

$y^{\left(4\right)}=m(m-1)(m-2)(m-3)x^{(m-4)}$ (5)

Substitute, equations (2), (3), (4), (5) in (1) and $y=x^m$

$$\begin{gathered} \left(x^4{x}^{(m-4)}\left[m\right(m-1\left)\right(m-2\left)\right(m-3\left)\right]+6x^3{x}^{(m-3)}\left[m\right(m-1\left)\right(m-2\left)\right] \\ +9x^2{x}^{(m-2)}\left[m\right(m-1\left)\right]+3x{x}^{(m-1)}\left[m\right]+x^m \\ \right)=0 \\ \left(x^m\left[m\right(m-1\left)\right(m-2\left)\right(m-3\left)\right]+6x^m\left[m\right(m-1\left)\right(m-2\left)\right]+9x^m\left[m\right(m-1\left)\right]+3x^m\left[m\right]+x^m\right)=0 \\ \left(x^m\left[m\right(m-1\left)\right(m-2\left)\right(m-3)+6m(m-1\left)\right(m-2)+9m(m-1)+3m+1]\right)=0 \\ \left(x^m\left[\left(m^2-m\right)\left(m^2-5m+6\right)+6m\left(m^2-3m+2\right)+\left(9m^2-9m\right)+(3m+1)\right]\right)=0 \\ \left(m^4+2m^2+1\right)x^m=0 \end{gathered}$$

Find the solution by obtaining the roots of an equation

Hence, from the above equation $x^m$cannot be equal to 0,

$$\begin{gathered} m^4+2m^2+1=0 \\ {\left(m^2+1\right)}^2=0 \end{gathered}$$

Thus, the roots are,$m_{1,2}=\pm iand{m}_{3,4}=\pm i$

Therefore, we can obtain the solution for the given equation as,

$\begin{array}{rcl}{y}_h& =& c_1{x}^i+c_2{x}^{-i}+c_3\ln x{x}^i+c_4\ln x{x}^{-i}\\ & =& c_1{\left(e^{\ln x}\right)}^i+c_2{\left(e^{\ln x}\right)}^{-i}+c_3\ln x{\left(e^{\ln x}\right)}^i+c_2\ln x{\left(e^{\ln x}\right)}^{-i}\\ & =& c_1\left(e^{i\ln x}\right)+c_2\left(e^{-i\ln x}\right)+c_3\ln x\left(e^{i\ln x}\right)+c_2\ln x\left(e^{-i\ln x}\right)\\ y& =& c_1\cos \left(\ln x\right)+c_2\sin \left(\ln x\right)+c_3\ln x\cos \left(\ln x\right)+c_4\ln x\sin \left(\ln x\right)\\ & & \end{array}$

Hence, the required solution for the given equation is. $y=c_1\text{cos}\left(\text{ln}x\right)+c_2\text{sin}\left(\text{ln}x\right)+c_3\text{ln}x\text{cos}\left(\text{ln}x\right)+c_4\text{ln}x\text{sin}\left(\text{ln}x\right)$

.