设p(χ),q(χ),f(χ)均是χ的连续函数,y1(χ),y2(χ),y3(χ)是y〞+p(χ)y′+q(χ)y=f(χ)的三个线性无关解,C1,C2为任意常数,则齐次方程y〞+p(χ)y′+q(χ)y=0的通解是( )

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问题 设p(χ),q(χ),f(χ)均是χ的连续函数,y1(χ),y2(χ),y3(χ)是y〞+p(χ)y′+q(χ)y=f(χ)的三个线性无关解,C1,C2为任意常数,则齐次方程y〞+p(χ)y′+q(χ)y=0的通解是(    )

选项 A、C1y1(χ)+(C1-C2)y2(χ)+(1-C2)y3(χ)
B、(C1-C2)y1(χ)+(C2-1)y2(χ)+(1-C1)y3(χ)
C、(C1+C2)y1(χ)+(C1-C2)y2(χ)+(1-C1)y3(χ)
D、C1y1(χ)+C2y2(χ)+(1-C1-C2)y3(χ)

答案B

解析 将选项B改写为:(C1-C2)y1(χ)+(C2-1)y2(χ)+(1-C1)y3(χ)
    =C1[y1(χ)-y3(χ)]+C2[y2(χ)-y1(χ)]+[y3(χ)-y2(χ)].
    因为y1(χ),y2(χ),y3(χ)均是y〞+p(χ)y′+q(χ)y=f(χ)的解,
    所以y1(χ)-y3(χ),y2(χ)-y1(χ),y3(χ)-y2(χ)均是y〞+p(χ)y′+q(χ)y=0的解,并且y1(χ)-y2(χ)与y2(χ)-y1(χ)线性无关.故B为通解.
    (事实上,若y1(χ)-y3(χ)与y2(χ)-y1(χ)线性相关,则存在不全为零的k1,k2使得
    k1[y1(χ)-y3(χ)]+k2[y2(χ)-y1(χ)]=0,
    即(k1-k2)y1(χ)+k2y2(χ)-k1y3(χ)=0.
    由于y1(χ),y2(χ),y3(χ)是线性无关的,故k1,k2全为零,矛盾.故y1(χ)-y3(χ)与y2(χ)-y1(χ)线性无关).
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