Find a general solution to the given differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{5}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{13}{\mathbf{x}}^{\mathbf{-}\mathbf{2}}\mathbf{y}\mathbf{,} \mathbf{x}\mathbf{>}\mathbf{0}$

The given differential equation is,

$\begin{array}{rcl}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}& \mathbf{=}& \mathbf{5}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{13}{\mathbf{x}}^{\mathbf{-}\mathbf{2}}\mathbf{y}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{13}{\mathbf{x}}^{\mathbf{-}\mathbf{2}}\mathbf{y}& \mathbf{=}& \mathbf{0}\\ {\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathbf{xy}\mathbf{\text{'}}\mathbf{+}\mathbf{13}\mathbf{y}& \mathbf{=}& \mathbf{0} .......\left(\mathbf{1}\right)\\ & & \end{array}$

Let,

$\begin{array}{rcl}\mathbf{x}& \mathbf{=}& {\mathbf{e}}^{\mathbf{t}}\\ \mathbf{dx}& \mathbf{=}& {\mathbf{e}}^{\mathbf{t}}\mathbf{dt}\\ \frac{\mathbf{dt}}{\mathbf{dx}}& \mathbf{=}& {\mathbf{e}}^{\mathbf{-}\mathbf{t}}\\ \mathbf{y}\mathbf{\text{'}}& \mathbf{=}& \frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{dy}}{\mathbf{dt}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}& \mathbf{=}& \frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\left(\frac{\mathbf{dy}}{\mathbf{dt}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\left(\mathbf{-}\mathbf{1}\right)\mathbf{+}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\right)\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\left(\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{dy}}{\mathbf{dt}}\right)\end{array}$

Substitute the value of $\mathbf{x}\mathbf{,} \mathbf{y}\mathbf{\text{'}}$and $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}$in the equation (1),

$\begin{array}{rcl}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathbf{xy}\mathbf{\text{'}}\mathbf{+}\mathbf{13}\mathbf{y}& \mathbf{=}& \mathbf{0}\\ {\mathbf{e}}^{\mathbf{2}\mathbf{t}}\left[{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\left(\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{dy}}{\mathbf{dt}}\right)\right]\mathbf{-}\mathbf{5}{\mathbf{e}}^{\mathbf{t}}\left[\frac{\mathbf{dy}}{\mathbf{dt}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\right]\mathbf{+}\mathbf{13}\mathbf{y}& \mathbf{=}& \mathbf{0}\\ \frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{-}\mathbf{5}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{+}\mathbf{13}\mathbf{y}& \mathbf{=}& \mathbf{0}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{13}\mathbf{y}& \mathbf{=}& \mathbf{0} ......\left(\mathbf{2}\right)\end{array}$

The auxiliary equation for the above equation,${\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{6}\mathbf{m}\mathbf{+}\mathbf{13}\mathbf{=}\mathbf{0}.$

Solve the auxiliary equation,

$\begin{array}{rcl}{\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{6}\mathbf{m}\mathbf{+}\mathbf{13}& \mathbf{=}& \mathbf{0}\\ \mathbf{m}& \mathbf{=}& \frac{\mathbf{6}\mathbf{\pm }\sqrt{\mathbf{36}\mathbf{-}\mathbf{52}}}{\mathbf{2}}\\ \mathbf{m}& \mathbf{=}& \frac{\mathbf{6}\mathbf{\pm }\sqrt{\mathbf{-}\mathbf{16}}}{\mathbf{2}}\\ \mathbf{m}& \mathbf{=}& \mathbf{3}\mathbf{\pm }\mathbf{2}\mathbf{i}\\ & & \end{array}$

The roots of the auxiliary equation are,${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{+}\mathbf{2}\mathbf{i}\mathbf{,} \& {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{-}\mathbf{2}\mathbf{i}.$

Therefore, the general solution is $\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{cos}\left(\mathbf{2}\mathbf{t}\right)\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{sin}\left(\mathbf{2}\mathbf{t}\right).$

Now substitute $t\mathbf{=}lnx$in the above equation,

$\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{x}}^{\mathbf{3}}\mathbf{cos}\left(\mathbf{2}\mathbf{lnx}\right)\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{x}}^{\mathbf{3}}\mathbf{sin}\left(\mathbf{2}\mathbf{lnx}\right)$

Thus, the general solution to the given differential equation is;

$\mathrm{y}={\mathrm{c}}_1{\mathrm{x}}^3\cos \left(2\mathrm{lnx}\right)+{\mathrm{c}}_2{\mathrm{x}}^3\sin \left(2\mathrm{lnx}\right)$