Answer
$y=e^{\frac{-2x}{3}}[1+\frac{2x}{3}]$
Work Step by Step
$9y''+12y'+4y=0$
Use auxiliary equation
$9r^{2}+12r+4=0$
$r=\frac{-12±\sqrt ((12)^{2}-4(9)(4))}{2(9)}$
$r=\frac{-12±\sqrt (144-144)}{18}$
$r=-\frac{2}{3}$
Formula 10
$y=c_{1}e^{rx}+c_{2}xe^{rx}$
$y=c_{1}e^{\frac{-2x}{3}}+c_{2}xe^{\frac{-2x}{3}}$
Given: $y(0)=1$
$y(0)=c_{1}e^{0}+c_{2}\times (0)e^{0}$
$1=c_{1}$
Given: $y'(0)=0$
$y'=-\frac{2}{3}c_{1}e^{\frac{-2x}{3}}+c_{2}[e^{\frac{-2x}{3}}-\frac{2x}{3}e^{\frac{-2x}{3}}]$
$y'(0)=-\frac{2}{3}c_{1}e^{0}+c_{2}[e^{0}-\frac{0}{3}e^{0}]$
Substitute
$0=-\frac{2}{3}(1)+c_{2}$
$c_{2}=\frac{2}{3}$
$y=e^{\frac{-2x}{3}}+\frac{2x}{3}e^{\frac{-2x}{3}}$
$y=e^{\frac{-2x}{3}}[1+\frac{2x}{3}]$
