(1)xy''+y'=0
两边积分,得
xy'=C
y'=C/x
y=C1ln|x|+C2
(2)令p=y'
则y''=dy'/dx=dp/dy * dy/dx=p' *p
于是方程化为pp'+p^2=1
p'=(1-p^2)/p
pdp/(1-p^2)=dy
1/2 * d(p^2)/(1-p^2)=dy
两边积分,得
-ln|1-p^2|=2y+C
-ln|1-y'^2|=2y+C
将y(0)=0,y'(0)=0代入,得C=0
于是方程化为 -ln|1-y'^2|=2y
ln|1-y'^2|= -2y
|1-y'^2|=e^(-2y)
1-y'^2=e^(-2y) 或 1-y'^2= -e^(-2y)(将y(0)=0,y'(0)=0代入,发现第二种情况不符,舍去)
接下去见图