Solve the following equations using method (d) above.

$x^2{y}^{\text{'}\text{'}}-3x{y}^{\text{'}}+4y=6x^2\ln x$

The given differential equation is $x^2{y}^{\text{'}\text{'}}-3x{y}^{\text{'}}+4y=6x^2\ln x$.

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.

$$\begin{gathered} x^2{y}^{\text{'}\text{'}}-3x{y}^{\text{'}}+4y=6x^2\ln x \\ \text{Let}x=e^z· \\ \text{So,}x=e^z \\ z=\ln x \end{gathered}$$

Solve further

$\frac{dz}{dx}=\frac{1}{x}$

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,

$\left(D\right(D-1)-3D+4)y=6e^{2z}z$

The auxiliary equation of the above equation is,

$m(m-1)-3m+4=0$

The solution of the auxiliary equation is,

$m=2,2$

Now,

$Q=6z{e}^{2z}$

So,

$$\begin{gathered} y_p=\frac{6z{e}^{2z}}{\left(D^2-4d+4\right)} \\ =\frac{6z{e}^{2z}}{\left({(D+2)}^2-4(D+2)+4\right)} \\ =\frac{6e^{2z}}{D^2}z \end{gathered}$$

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

$$\begin{gathered} y_{\text{p}}=\frac{6e^{2z}}{D^2}z \\ =\frac{6e^{2z}}{D}\left(\frac{z^2}{2}\right) \\ =z^3{e}^{2z} \end{gathered}$$

Thus, the complete solution is given as,

$$\begin{gathered} y=y_{\text{c}}+y_{\text{p}} \\ =\left(c_{1}z+c_2\right)e^{2z}+z^3{e}^{2z} \\ =\left(c_1\ln x+c_2\right)x^2+(\ln x{)}^3{x}^2 \end{gathered}$$

Therefore, the general solution of the equation is$y=\left(c_1\ln x+c_2\right)x^2+(\ln x{)}^3{x}^2$