• 2022-06-06
    设线性无关的函数y1(x),y2(x),y3(x)均是方程yˊˊ+p(x)yˊ+q(x)y=f(x)的解C1,C2是任意常数,则该方程的通解是 ( )
    A: C1y1+C2y2+y3
    B: C1y1+C2y2-(C1+C2)y3
    C: C1y1+C2y2-(1-C1-C2)y3
    D: C1y1+C2y2+(1-C1-C2)y3