• 2022-06-06
    设函数y1(x),y2(x)是微分方程y"+p(x)y=q(x)的两个不同特解,则该方程的通解为______
    A: y=C1y1+C2y2(C1,C2为任意常数).
    B: y=y1+Cy2(C为任意常数).
    C: y=y1+C(y1+y2)(C为任意常数).
    D: y=y1+C(y2-y1)(C为任意常数).