微分方程通解的问题,解释一下为什
y'+p(x)y=e(x)
因为y1(x)、y2(x)为它的解,所以有
(y1)'+p(x)y1=e(x)
(y2)'+p(x)y2=e(x)
A。把c(y1-y2)代入微分方程的左边,得
(c(y1-y2))'+p(x)(c(y1-y2))
=c((y1)'-(y2)')+p(x)(cy1-cy2)
=c((y1)'+p(x)y1)-c((y2)'+p(x)y2)
=ce(x)-ce(x)=0
所以c(y1-y2)不是方程的解。
B。把y1+c(y1-y2)代入微分方程的左边,得
(y1+c(y1-y2))'+p(x)(y1+c(y1-y2))
=(y1)'+c(y1-y2)'+p(...全部
y'+p(x)y=e(x)
因为y1(x)、y2(x)为它的解,所以有
(y1)'+p(x)y1=e(x)
(y2)'+p(x)y2=e(x)
A。把c(y1-y2)代入微分方程的左边,得
(c(y1-y2))'+p(x)(c(y1-y2))
=c((y1)'-(y2)')+p(x)(cy1-cy2)
=c((y1)'+p(x)y1)-c((y2)'+p(x)y2)
=ce(x)-ce(x)=0
所以c(y1-y2)不是方程的解。
B。把y1+c(y1-y2)代入微分方程的左边,得
(y1+c(y1-y2))'+p(x)(y1+c(y1-y2))
=(y1)'+c(y1-y2)'+p(x)y1+cp(x)(y1-y2)
=(y1)'+p(x)y1+c((y1)'+p(x)y1)-c((y2)'+p(x)y2)
=e(x)+ce(x)-ce(x)
=e(x)
因此,y1+c(y1-y2)是微分方程的解。
用同样的方法,可以得出C、D都不是方程的解。所以选B。收起