The differential equation

$x^2\frac{d^{2}y}{d{x}^2}-7x\frac{dy}{dx}+16y=0$

has $x^4$ as a solution.

Applying reduction order we set $y_2=u{x}^4.$

Then $($using the prime notation for the derivatives$)$

$y_2^{\text{'}}=u^{\text{'}}{x}^4+4u{x}^3$

$y_2^{\text{'}\text{'}}=u^{\text{'}\text{'}}{x}^4+4u^{\text{'}}{x}^3+12u{x}^2$

So$,$ plugging $y_2$ into the left side of the differential equation, and reducing, we get

$x^2{y}_2^{\text{'}\text{'}}-7x{y}_2^{\text{'}}+16y_2=x^6{u}^{\text{'}\text{'}}-3x^5{u}^{\text{'}}$

The reduced form has a common factor of $x^5$ which we can divide out of the equation so that we have $x{u}^{\text{'}\text{'}}+u^{\text{'}}=0.$

Since this equation does not have any u terms in it we can make the substitution $w=u^{\text{'}}$ giving us the first order linear equation $x{w}^{\text{'}}+w=0.$

This equation has integrating factor $x^6{u}^{\text{'}\text{'}}-3x^5{u}^{\text{'}},$ for $x>0.$

If we use a as the constant of integration, the solution to this equation is $w=$

$$

Integrating to get u $,$ and using b as our second constant of integration we have $u=$

$(\frac{a}{4})x^4+b$

Finally $y_2=(\frac{a}{4})x^8+b{x}^4$ and the general solution is