Question: In Problems 15to 18, 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 in Problem 14b.

$\left(x^2+1\right){\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}{\left(x^2+1\right)}^{\mathbf{2}}\mathbf{;}\mathit{x}\mathbf{,}\mathbf{1}\mathbf{-}{\mathit{x}}^{\mathbf{2}}$

The given parameters are$x,1-x^2$

Homogeneous equation: A homogeneous function of x and y of degree nmeans a function can be written as${\mathit{x}}^{\mathbf{n}}\mathit{f}\left(\frac{y}{x}\right)$

$\mathit{P}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}\mathit{d}\mathit{x}\mathbf{+}\mathit{Q}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

Where Pand Qare homogeneous function of the same degree is called homogeneous.

The impulse response of an inhomogeneous linear differential operator defined on a domain with specified initial or boundary conditions is known as a green’s function.

The given equation is.

$\begin{array}{c}{y}^{\text{'}\text{'}}-\frac{2x}{\left(x^2+1\right)}\cdot y^{\text{'}}+\frac{2}{\left(x^2+1\right)}y=\left(x^2+1\right)\\ y^{\text{'}\text{'}}+p\left(x\right)y^{\text{'}}+Q\left(x\right)y=R\left(x\right)\end{array}$

The parameters $x,1-x^2$are solutions of homogeneous equations.

$\begin{array}{l}{y}_c=C_{1}x+C_{2}1-x^2\\ y_c=C_{1}u+C_{2}v\end{array}$

The particular solution is.

$y_p=Au+Bv$

The value of coefficient A is

$\begin{array}{l}A=-\int \frac{vR}{u{v}^{\text{'}}-v{u}^{\text{'}}}dx\\ A=-\int \frac{\left(1-x^2\right)\left(1+x^2\right)}{\left(x\right){\left(1-x^2\right)}^{\text{'}}-{\left(x\right)}^{\text{'}}\left(1-x^2\right)}dx\\ A=-\int \frac{\left(1-x^2\right)\left(1+x^2\right)}{\left(x\right)(0-2x)-1.\left(1-x^2\right)}dx\\ A=-\int \frac{\left(1-x^2\right)\left(1+x^2\right)}{-2x^2-1+x^2}dx\end{array}$

Solve the equation further

$\begin{array}{l}A=-\int \left(1-x^2\right)dx\\ A=x-\frac{x^3}{3}\end{array}$

The value of coefficient B is

$\begin{array}{l}B=-\int \frac{uR}{u{v}^{\text{'}}-v{u}^{\text{'}}}dx\\ B=-\int \frac{x\left(1+x^2\right)}{-\left(1+x^2\right)}dx\\ B=-\int -xdx\\ B=-\frac{x^2}{2}\end{array}$

Substitute the value in a particular solution.

$\begin{array}{l}{y}_p=Au+Bv\\ y_p=\left(x-\frac{x^3}{3}\right)x-\frac{x^2}{2}\left(1-x^2\right)\\ y_p=x^2-\frac{x^4}{3}-\frac{x^2}{2}+\frac{x^4}{2}\\ y_p=\frac{x^2}{2}+\frac{1}{6}{x}^4\end{array}$

The value of the general equation is.

$\begin{array}{l}y=y_c+y_p\\ y=C_{1}x+C_2\left(1-x^2\right)+\frac{x^2}{2}+\frac{1}{6}{x}^4\end{array}$

Thus, the general solution of is _{$\left(x^2+1\right)y^{\text{'}\text{'}}-2x{y}^{\text{'}}+2y={\left(x^2+1\right)}^2$}

_{$y=C_{1}x+C_2\left(1-x^2\right)+\frac{x^2}{2}+\frac{1}{6}{x}^4$}

and particular solution is $y_p=\frac{x^2}{2}+\frac{1}{6}{x}^4$.