第六章常微分方程 例1:xy”+xy2-y=0 解:可观察出一个解:y1=x 今利用侍定函数法求第二个特解(变动任意常数法) x)×y2=(x) p(×y2+c(m+c()y(x) 1×y2=(x)+2c(x)(x)+c(x)y(x) 0=4(x)0+c(py+2y1)+c(x)(x) c"(x) (x)p(x)y1(x)+2y(x) hnc(x) (x) d x p(x)y1(x)+2y(x) -n(x) d(x)=e p(an(2yah dx 解是y=cx+c2x2+2x+2) 例2: +2xy-2y=0 解:多项式解:y=x",n-3n+2=0 1=x;y2=x2+1 例3:(1-x2)y”+2xy-2y=1-x2, 9x)xr=c(x)y(x)+c2(x)y2(x) p(x)×H=c{(xy(x)+c(x)y(x)+c(x)n(x)+c2()2(x) 1×y1=c(x)(x)+c2(x)2(x)+cf(xy(x)+c(n(x) (x)=c(x)0+c(x)0+c(x)y(x)+c()() 第六章常微分方程第六章 常微分方程 第六章 常微分方程 例 1： xy  + xy  − y = 0 解：可观察出一个解： y = x 1 今利用侍定函数法求第二个特解 ( 变动任意常数法): ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 p x y x 2y x y x c x c x +  − =   ( ) ( ) ( ) ( ) ( )  +  −  = dx p x y x y x y x c x 1 1 1 2 ln( ) ( ) ( ) ( ) ( ) c x e dx dx p x y x y x y x   +  − = 1 1 1 2 解是 ( ) x y c x c x x x e − = + + 2 + 2 2 1 2 . 例 2： (1 ) 2 2 0 2 − x y  + xy  − y = 解：多项式解： n y = x , 3 2 0 2 n − n + = ; y = x 1 ; 1 2 y2 = x + . 例 3： ( ) 2 2 1− x y  + 2xy  − 2y =1− x , q(x) ( ) 2 1 y = c x y p(x) y c(x)y c (x)y (x) 2 1 1  =  +  1 y c(x)y c (x)y (x) c (x)y (x) 2 1 2 1 1  = +   +  c(x) c (p y y ) c (x)y (x) 0 0 1 2 1 1 =  +  +  +  q(x) Y c (x)y (x) c (x)y (x) = 1 1 + 2 2 O p(x) Y c (x)y (x) c (x)y (x) c (x)y (x) c (x)y (x) 2 1 1 1 1 1 1 2 2  =  +  +  +  1 Y c (x)y (x) c (x)y (x) c (x)y (x) c (x)y (x) 1 1 2 2 1 1 2 2  =  +  +   +   f (x) c(x) c(x) c (x)y (x) c (x)y (x) 0 0 1 1 2 2 =  +  +   +  