Find a general solution to the Cauchy-Euler equation ${\mathit{x}}^{\mathbf{3}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{3}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{4}}\mathit{c}\mathit{o}\mathit{s}\mathit{x}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{x}\mathbf{>}\mathbf{0}$

Variation of parameters, also known as variation of constants, is a general method to solve inhomogeneous linear ordinary differential equations.

Consider the following Cauchy-Euler equation$x^3{y}^{\text{'}\text{'}\text{'}}-3x{y}^{\text{'}}+3y=x^4\cos x$

Let$y=x^r$ be the solution of homogeneous equation$x^3{y}^{\text{'}\text{'}\text{'}}-3x{y}^{\text{'}}+3y=0$

Then,

$$\begin{gathered} y^{\text{'}}=r{x}^{r-1} \\ {y}^{\text{'}\text{'}}=r(r-1)x^{r-2} \\ {y}^{\text{'}\text{'}\text{'}}=r(r-1)(r-2)x^{r-3} \end{gathered}$$

Substitute the above values in $x^3{y}^{\text{'}\text{'}\text{'}}-3x{y}^{\text{'}}+3y=0$.

$r^3-3r^2-r+3=0$

So, all the roots of the equation are $r=1,-1,3$.

Thus, the solutions of the homogenous equation are$=\left\{x,x^{-1},x^3\right\}$

So, the general solution of homogenous equation is $y=C_{1}x+C_2{x}^{-1}+C_3{x}^3$.

$$\begin{gathered} W_1\left(x\right)=(-1{)}^{3-1}W\left[y_2,y_3\right] \\ =\left(1\right)\left|\begin{array}{cc}{x}^{-1}& x^3\\ -x^2& 3x^2\end{array}\right| \\ =4x \\ {W}_2=(-1{)}^{3-2}W\left[x{x}^3\right] \\ =-2x^3 \\ {W}_3=(-1{)}^{3-3}W\left[x{x}^{-1}\right] \\ =-2x^{-1} \end{gathered}$$Wronskian of$y_1,y_2,y_3$ is,

$$\begin{gathered} W\left[y_1,y_2,y_3\right]=\left|\begin{array}{ccc}{y}_1& y_2& y_3\\ y{\text{'}}_1& y{\text{'}}_2& y{\text{'}}_3\\ y{\text{'}\text{'}}_1& y{\text{'}\text{'}}_2& y{\text{'}\text{'}}_3\end{array}\right| \\ W\left[x_1,x_2,x_3\right]=\left|\begin{array}{ccc}x& x^{-1}& x^3\\ 1& -x^{-2}& 3x^2\\ 0& 2x^{-3}& 6x\end{array}\right| \\ =-16 \end{gathered}$$

We know that$v_k\left(x\right)=\int \frac{g\left(x\right)W_k\left(x\right)}{W\left[y_1,y_2,y_3\right]}dx$

Hence,

$$\begin{gathered} v_1\left(x\right)=\int \frac{g\left(x\right)W_1\left(x\right)}{W\left[y_1,y_2,y_3\right]}dx \\ =\int \frac{\left(x\cos x\right)\left(4x\right)}{-16}dx \\ =-\frac{1}{4}\left[x^2\sin x+2x\cos x-2\sin x\right] \end{gathered}$$

$$\begin{gathered} v_2\left(x\right)=\int \frac{g\left(x\right)W_2\left(x\right)}{W\left[y_1,y_2,y_3\right]}dx \\ =\int \frac{\left(x\cos x\right)\left(-2x^3\right)}{-16}dx \\ =\frac{1}{8}\left[x^4\sin x+4x^3\cos x-12x^2\sin x-24x\cos x+24\sin x\right] \end{gathered}$$

And

$$\begin{gathered} v_3=\int \frac{g\left(x\right)W_3\left(x\right)}{W\left[y_1,y_2,y_3\right]}dx \\ =\frac{1}{8}\sin x \end{gathered}$$

Substitute the values$v_1,v_2,v_3$in $y_p$.

$$\begin{gathered} y_p=v_{1}x+v_2{x}^{-1}+v_3{x}^3 \\ =-\frac{1}{4}\left[x^2\sin x+2x\cos x-2\sin x\right]x+\frac{1}{8}\left[x^4\sin x+4x^3\cos x-12x^2\sin x-24x\cos x+24\sin x\right]x^{-1}+\left[\frac{1}{8}\sin x\right]x^3 \\ =-\frac{1}{4}\left[x^3\sin x+2x^2\cos x-2x\sin x\right]+\frac{1}{8}\left[x^3\sin x+4x^2\cos x-12x\sin x-24\cos x+24x^{-1}\sin x\right]+\left[\frac{1}{8}{x}^3\sin x\right] \end{gathered}$$

Thus, the particular-solution is.$y_p=-x\sin x-3\cos x+3x^{-1}\sin x$

Thus, the general solution is:$y=C_{1}x+C_2{x}^{-1}+C_3{x}^3-x\sin x-3\cos x+\frac{3\sin x}{x}.$