单选题
设函数p(x)，q(x)，f(x)连续，而y1，y2，y3都是方程y"+p(x)y'+q(x)y=f(x)的解，则该方程必定还有解
【正确答案】
A
【答案解析】[分析] 选项(A)正确．
事实上，
(y1+y2-y3)"+p(x)(y1+y2一y3)'+q(x)(y1+y2-y3)
=(y"1+p(x)y'1+q(x)y1)+(y"2+p(x)y'2+q(x)y2)-(y"3+p(x)y'3+q(x)y3)
=f1(x)+f2(x)-f3(x)
=f(x)．
故y1+y2-y3也为方程y"+p(x)y'+q(x)y=f(x)的解．