Question: Use the given solutions of the homogeneous equation to find a particular solution of the given equation. You can do this either by the Green function formulas in the text or by the method of variation of parameters

${\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathit{x}\mathit{l}\mathit{n}\mathit{x}\mathbf{;}\mathit{x}\mathbf{,}{\mathit{x}}^{\mathbf{2}}$

The given expressions are $x^2{y}^{\text{'}\text{'}}-2x{y}^{\text{'}}+2y^{\text{'}\text{'}}=x\ln x$.

In mathematics, a Green's function is the impulse response of an inhomogeneous linear differential operator defined on a domain with specified initial conditions or boundary conditions.

The complementary function of given equation.

$y_c=C_1{y}_1+C_2{y}_2$

The particular solution is.

$y_p=A{y}_1+B{y}_2$

The coefficient Aand Bare,

$\begin{array}{c}A=-\int \frac{y_{1}R\left(x\right)}{W\left(y_1,y_2\right)}dx\\ B=\int \frac{y_{1}R\left(x\right)}{W\left(y_1,y_2\right)}dx\\ \text{Here,}y\text{1}=x^2\text{and}{y}_2=xW\\ =\left|\begin{array}{ll}{x}^2& x\\ 2x& 1\end{array}\right|\\ =-x^2\end{array}$

Substitute value in complementary function

$y=C_1{x}^2+C_{2}x$

Substitute value in the particular solution is.

$y_p=A{x}^2+Bx$

The value of coefficient A is

$\begin{array}{c}A=-\int \frac{yR\left(x\right)}{W\left(y_1,y_2\right)}dx\\ A=-\int \frac{x\left(\frac{\ln x}{x}\right)}{-x^2}dx\\ A=-\frac{h+1}{x}\end{array}$

The value of coefficient B is,

$\begin{array}{c}B=\int \frac{yR\left(x\right)}{W(y,y)}dx\\ B=-\int \frac{x^2\left(\frac{\ln x}{x}\right)}{-x^2}dx\\ B=-\frac{{\left(\ln x\right)}^2}{2}\end{array}$

Substitute the value in particular solution.

$$\begin{gathered} y_p=A{y}_1+B{y}_2. \\ \begin{array}{c}{y}_p=-x^2\left(\frac{\ln x+1}{x}\right)-x\left(\frac{\ln x)}{2}\right)\\ y_p=-x\left(\ln x+1+\frac{{\left(\ln x\right)}^2}{2}\right)\end{array} \end{gathered}$$

The value of general equation is,

$y=C_{1y}+C_{2y2}-\frac{x}{2}{\left(\ln x\right)}^2-x\left(\ln x\right)-x$

Thus, the value of$x^2{y}^{\text{'}}-2x{y}^{\text{'}}+2y^{\text{'}}=x\ln x$ for general solution is$y_c=C_1{y}_1+C_2{y}_2-\frac{x}{2}{\left(\ln x\right)}^2-x\left(\ln x\right)-x$ and particular solution is$yp=-x\left(\ln x+1+\frac{{\left(\ln x\right)}^2}{2}\right)$