Okay,

You know that,

(y1)''+p(x)(y1)'+q(x)(y1)=r1(x)

(y2)''+p(x)(y2)'+q(x)(y2)=r2(x)

Then add your equations,

[(y1)''+(y2)'']+[p(x)(y1)'+p(x)(y1)']+[q(x)(y1)+q(x)(y2)]=r1(x)+r2(x)

Factor,

[(y1)''+(y2)'']+p(x)[(y1)'+(y2)']+q(x)[(y1)+(y2)]=r1(x)+r2(x)

Use the addition property of the derivative,

(Note if you want to sound cool say diffrenciation is a homomorphism)

(y1+y2)''+p(x)(y1+y2)'+q(x)(y1+y2)=r1(x)+r2(x)

Therefore we see that,

y=y1+y2

Is the solution.