Make the substitution
, then compute the derivatives of
with respect to
via the chain rule.
![(\mathrm dy)/(\mathrm dx)=(\mathrm dy)/(\mathrm dt)(\mathrm dt)/(\mathrm dx)](https://img.qammunity.org/2020/formulas/mathematics/college/1ez2hgji7bdpg40resut7mn6v38eu7tqiz.png)
![\implies(\mathrm dy)/(\mathrm dx)=\frac1x(\mathrm dy)/(\mathrm dt)](https://img.qammunity.org/2020/formulas/mathematics/college/v2t44ykf1qszubp2fdew7blyq970q4bdiq.png)
Let
.
![(\mathrm d^2y)/(\mathrm dx^2)=(\mathrm d)/(\mathrm dx)\left[\frac fx\right]=(x(\mathrm df)/(\mathrm dx)-f)/(x^2)](https://img.qammunity.org/2020/formulas/mathematics/college/setkbry7hvvgv3gj4zl72vhhuafam1rjr4.png)
![(\mathrm df)/(\mathrm dx)=(\mathrm df)/(\mathrm dt)(\mathrm dt)/(\mathrm dx)=\frac1x(\mathrm d^2y)/(\mathrm dt^2)](https://img.qammunity.org/2020/formulas/mathematics/college/e6toens2ws1dchufvlthhc9337gnhp72rk.png)
![\implies(\mathrm d^2y)/(\mathrm dx^2)=\frac1{x^2}\left((\mathrm d^2y)/(\mathrm dt^2)-(\mathrm dy)/(\mathrm dt)\right)](https://img.qammunity.org/2020/formulas/mathematics/college/kmuc8s8nl9qlmm1allgamol5ddfr2vi17z.png)
Let
.
![(\mathrm d^3y)/(\mathrm dx^3)=(\mathrm d)/(\mathrm dx)\left[(g-f)/(x^2)\right]=(x^2\left((\mathrm dg)/(\mathrm dx)-(\mathrm df)/(\mathrm dx)\right)-2x(g-f))/(x^4)](https://img.qammunity.org/2020/formulas/mathematics/college/anbkgyphunzwiwl7yng57pli7ff5b393dd.png)
![(\mathrm dg)/(\mathrm dx)=(\mathrm dg)/(\mathrm dt)(\mathrm dt)/(\mathrm dx)=\frac1x(\mathrm d^3y)/(\mathrm dt^3)](https://img.qammunity.org/2020/formulas/mathematics/college/4l45n9ba2n19gzf0rod0vce8oncwtxb83g.png)
![\implies(\mathrm d^3y)/(\mathrm dx^3)=(x^2\left(\frac1x(\mathrm dg)/(\mathrm dt)-\frac1x(\mathrm df)/(\mathrm dt)\right)-2x(g-f))/(x^4)=\frac1{x^3}\left((\mathrm d^3y)/(\mathrm dt^3)-3(\mathrm d^2y)/(\mathrm dt^2)+2(\mathrm dy)/(\mathrm dt)\right)](https://img.qammunity.org/2020/formulas/mathematics/college/im5u5rwx4kp2zqyuo72p19wihsabj7ff8r.png)
Substituting
and its derivatives into the ODE gives a new one that is linear in
:
![\left((\mathrm d^3y)/(\mathrm dt^3)-3(\mathrm d^2y)/(\mathrm dt^2)+2(\mathrm dy)/(\mathrm dt)\right)+5\left((\mathrm d^2y)/(\mathrm dt^2)-(\mathrm dy)/(\mathrm dt)\right)+7(\mathrm dy)/(\mathrm dt)+8y=0](https://img.qammunity.org/2020/formulas/mathematics/college/er12ejfodcj8qjfjl3m4armpz9paezc68t.png)
![(\mathrm d^3y)/(\mathrm dt^3)+2(\mathrm d^2y)/(\mathmr dt^2)+4(\mathrm dy)/(\mathrm dt)+8y=0](https://img.qammunity.org/2020/formulas/mathematics/college/2644qxnrhf7qosi6qrvzy23b733n9rsmff.png)
![y'''+2y''+4y'+8y=0](https://img.qammunity.org/2020/formulas/mathematics/college/vti9m4569g5xdyojhp0f5r901b82npcu59.png)
which has characteristic equation
![r^3+2r^2+4r+8=(r+2)(r^2+4)=0](https://img.qammunity.org/2020/formulas/mathematics/college/xfw0bqqrjsr2webxt4kaoxr7uzu0yvg9wk.png)
with roots
and
, so that the characteristic solution is
![y_c(t)=C_1e^(-2t)+C_2\cos2t+C_3\sin2t](https://img.qammunity.org/2020/formulas/mathematics/college/sksh8vnuc9aoyai8uv6kvpdsrqo5sy1mtp.png)
Replace
to solve for
:
![y_c(x)=C_1e^(-2\ln x)+C_2\cos(2\ln x)+C_3\sin(2\ln x)](https://img.qammunity.org/2020/formulas/mathematics/college/bbvo10969vhv55j1gov0igprhw6s7eznwu.png)
![\boxed{y(x)=(C_1)/(x^2)+C_2\cos(2\ln x)+C_3\sin(2\ln x)}](https://img.qammunity.org/2020/formulas/mathematics/college/r0kdq1gd9oit2etgt15l5dt4dbvczf6i9z.png)