Answer
$y=\frac{-27}{10}sin(\frac{1}{3}x)+\frac{9}{10}cos(\frac{1}{3}x)+3x+\frac{1}{10}e^{-x}$
Work Step by Step
$y''+y=3x+e^{-x}$
General solution: $y=y_c+y_p=c_1sin(\frac{1}{3}x)+c_2cos(\frac{1}{3}x)+3x+\frac{1}{10}e^{-x}$
Plug in the given values: $y(0)=1$ and $y'(0)=2$
Hence, $y=\frac{-27}{10}sin(\frac{1}{3}x)+\frac{9}{10}cos(\frac{1}{3}x)+3x+\frac{1}{10}e^{-x}$
