命题1初值问题(31)等价于积分方程d f(xy)(3.1) y(x0) y=y0+.f(y)dt(35) 证明:若y=q(x)为(3.1)的连续解,则 do(x) f(x,0(x) dx (x)=y 对第一式从x到x取定积分得 0(x)-0(x0)=「f(x,0(x)dx 即(x)=y0+.f(x,0(x)x 故y=0(x)为3.5的连续解命题1 初值问题(3.1)等价于积分方程 ( , ) (3.5) 0 y y0 f t y dt x x = + 证明: 若y =(x)为(3.1)的连续解,则 , ( ) ( , ( )) ( ) 0 0     = = x y f x x dx d x    对第一式从x0 到x取定积分得x x f x x dx x x − = 0 ( ) ( ) ( , ( ))   0  即 x y f x x dx x x = + 0 ( ) ( , ( ))  0  故y =(x)为(3.5)的连续解. ,(3.1) ( ) ( , ) 0 0     = = y x y f x y dx dy