Tutorial Exercise

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

$9y^{\text{'}\text{'}}-24y^{\text{'}}+16y=0$;$,y_1=e^{4\frac{x}{3}}$

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 $y_2(x).$

Step $1$

$We$ are given one solution $y_{1}of$ the following differential equation.

$9y^{\text{'}\text{'}}-24y^{\text{'}}+16y=0$;$,y_1=e^{4\frac{x}{3}}$

$To$ find a second solution $y_2$ using the reduction $of$ order formula

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

$we$ must write the given differential equation $in$ standard form.

$9y^{\text{'}\text{'}}-24y^{\text{'}}+16y=0$

$y^{\text{'}\text{'}}+P(x)y^{\text{'}}+Q(x)y=0$

$y^{\text{'}\text{'}}+(,)y^{\text{'}}+{(}^{\text{'}})y=0$