$y=y_1(t)+y_2(t)$ is solution of $y'+p(t)y=g(t)$ if $y=y_1(t)$ is one of $y'+p(t)y=0$ and $y=y_2(t)$ is one of $y'+p(t)y=g(t)$

Let $y=y_1(t)$ be a solution of $y'+p(t)y=0$ and let $y=y_2(t)$ be a solution of $y'+p(t)y=g(t)$. How can we show that $y=y_1(t)+y_2(t)$ is also a solution of $y'+p(t)y=g(t)$?

I'm not really sure how to approach the problem. It's in the section dealing with differences between linear and nonlinear equations.

-

$(y_1+y_2)'(t)+p(t)(y_1+y_2)(t)$
$=y_1'(t)+y_2'(t)+p(t)y_1(t)+p(t)y_2(t)$
$=y_1'(t)+p(t)y_1(t)+y_2'(t)+p(t)y_2(t)$
$=0+g(t)$
$=g(t)$