Answer
True
Work Step by Step
Since $y_1$ and $y_2$ are both solutions, and since the equation is homogenous, plugging $y_1$ and $y_2$ into the differential equation yields 0 both times. Now since differentiation is a linear operator, we have the following when plugging in $y_1$ + $y_2$:
($y_1$ + $y_2$)'' + ($y_1$ + $y_2$) = $y_1$'' + $y_2$'' + $y_1$ + $y_2$
= ($y_1$'' + $y_1$) + ($y_2$'' + $y_2$)
= 0 + 0
The last step follows from the fact that $y_1$ and $y_2$ are both solutions to the differential equation.
