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
As we are given that $y''+y=3x+e^{-x}$
The general solution is of the form $y=y_c+y_p=c_1sin(\frac{1}{3}x)+c_2cos(\frac{1}{3}x)+3x+\frac{1}{10}e^{-x}$
As per question, $y(0)=1$ and $y'(0)=2$ after plugging in the values, we have
$$y=\frac{-27}{10}sin(\frac{1}{3}x)+\frac{9}{10}cos(\frac{1}{3}x)+3x+\frac{1}{10}(e^{-x})$$
