Solve the following equations using method (d) above.

$x^2{y}^{\text{'}\text{'}}+x{y}^{\text{'}}-y=x-x^{-1}$

The given differential equation is $x^2{y}^{\text{'}\text{'}}+x{y}^{\text{'}}-y=x-x^{-1}$.

Auxiliary equation is an algebraic equation of degree $n$upon which depends the solution of a given$n^{th}$ -order differential equation or difference equation.

Consider the given equation.

$x^2{y}^{\text{'}\text{'}}+3x{y}^{\text{'}}-3y=0$

Let$x=e^z·\text{So}$ ,

$$\begin{gathered} x=e^z \\ z=\ln x \\ \frac{dz}{dx}=\frac{1}{x} \end{gathered}$$

Now,

$$\begin{gathered} \frac{dy}{dx}=\frac{dy}{dz}·\frac{dz}{dx} \\ =\frac{dy}{dz}\left(\frac{1}{x}\right) \\ \frac{d^{2}y}{d{x}^2}=\frac{1}{x^2}\left(\frac{d^{2}y}{d{z}^2}-\frac{dy}{dz}\right) \\ {x}^2\frac{d^{2}y}{d{x}^2}=\frac{d^{2}y}{d{z}^2}-\frac{dy}{dz} \end{gathered}$$

And,

$x\frac{dy}{dx}=\frac{dy}{dz}$

The given differential equation can be written as,

$\left(D\right(D-1)+D-1)y=e^z-e^{-z}$

The auxiliary equation of the above equation is,

$m(m-1)+m-1=0$

The solution of the auxiliary equation is,

$m=\pm 1$

Let the roots be represented as,

$$\begin{gathered} a=1 \\ b=-1 \end{gathered}$$

Now,

$$\begin{gathered} Q=e^z-e^{-z} \\ =Q_1-Q_2 \end{gathered}$$

Thus,

$y_{\text{p}}=y_1+y_2$

So,

$$\begin{gathered} y_1=\frac{e^z}{\left(D^2-1\right)} \\ =\frac{e^z}{\left(D^2+2D\right)} \\ =\frac{e^z}{2D\left(1+\frac{D}{2}\right)} \\ =\frac{e^z}{2D}{\left(1+\frac{D}{2}\right)}^{-1} \end{gathered}$$

Now, $D$represent the first derivative and$D^2$ represent the second derivative.

So,

$$\begin{gathered} y_1=\frac{e^z}{2D} \\ =\frac{z{e}^z}{2} \end{gathered}$$

And,

$y_2=-\frac{z{e}^{-z}}{2}$

Thus,

$y_{\text{p}}=\frac{z{e}^z}{2}-\frac{z{e}^{-z}}{2}$

Thus, the complete solution is given as,

$$\begin{gathered} y=y_{\text{c}}+y_{\text{p}} \\ =c_1{e}^{-z}+c_2{e}^z+\frac{z{e}^z}{2}-\frac{z{e}^{-z}}{2} \\ =c_1{x}^{-1}+c_{2}x+\frac{\left(\ln x\right)x}{2}-\frac{\left(\ln x\right)x^{-1}}{2} \\ =c_1{x}^{-1}+c_{2}x+\frac{1}{2}\left(x-x^{-1}\right)\ln x \end{gathered}$$

Therefore, the general solution of the equation is$y=c_1{x}^{-1}+c_{2}x+\frac{1}{2}\left(x-x^{-1}\right)\ln x$ .