Example:(X^2D^2-xD-15)y=0,y(1)=0.1,y'(1)=-4.5

Here the given differential equation is
${\displaystyle x^{2}y{}^{″}-x{y}^{\prime }-15y=0,y\left(0\right)=0.1,y^{\prime }\left(0\right)=-4.5}$
Let ${\displaystyle \ln x=z}$ then
${\displaystyle \frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}=\frac{1}{x}\frac{dy}{dz}}$.
similary
${\displaystyle x^2\frac{d^{2}y}{{dx}^2}=\frac{d^{2}y}{{dz}^2}-\frac{dy}{dz}}$.
Using above two expressions, the original differential equation reduces to
${\displaystyle (\frac{d^{2}y}{{dz}^2}-\frac{dy}{dz})-\frac{dy}{dz}-15y=0}$
${\displaystyle \Rightarrow \frac{d^{2}y}{{dz}^2}-2\frac{dy}{dz}-15y=0}$.
The auxiliary equation of the differential equation ${\displaystyle \frac{d^{2}y}{{dz}^2}-2\frac{dy}{dz}-15y=0}$ becomes
${\displaystyle m^2-2m-15=0}$
${\displaystyle \Rightarrow (m-5)(m+3)=0}$
${\displaystyle \Rightarrow m=5,-3}$.
Therefore the general solution of the differential equation is
${\displaystyle y\left(x\right)=a{e}^{5z}+b^{-3z}=a{x}^5+\frac{b}{x^3}}$.
Here we can not put y(0)=0.1 since in the general solution term ${\displaystyle \frac{b}{x^3}}$ is present.