The indicated function $y_1(x)is$ a solution $of$ the following differential equation.

$y^{\text{'}\text{'}}-8y^{\text{'}}+16y=0$;$,y_1=e^{4x}$

Use reduction $of$ order, $or$ the formula

$y_2=y_1(x){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{e^{-{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}P(x)dx}}{y_1^2(x)}dx$

$as$ instructed, $to$ find a second solution $y_2(x).$