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

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