C(x)y1(x)+C(x)1(x) +p(ac(r)y(x)=q(x) y(x)是Q的解, C(x)y,(x)+p(x)c(x)y(r)=0 化简得到C(x)(x)=q(x) 即 C(x)=g(xeJp()de 2021/2/202021/2/20 9 y1 ( x ) 是 (2)的解, ( ) '( ) ( ) ( ) ( ) 0 C x y1 x + p x C x y1 x = ( ) ( ) ( ) '( ) 1 1 C x y x + C x y x ( ) ( ) ( ) ( ) + p x C x y1 x = q x 化简得到 ( ) ( ) ( ) 1 C x y x = q x = p x dx C x q x e ( ) 即 ( ) ( )