y”=(y’/x)/(1+ln(y'/x)) y'(1)=1 y(1)=1/2

问题描述:

y”=(y’/x)/(1+ln(y'/x)) y'(1)=1 y(1)=1/2

p=y',p'(x+p)=p(dp/dx)(x+p)=p1.p≠0时 (dx/dp)=x/p+p令z=x/p,x=zp,x'=z+z'pz+z'p=z+pz'p=px/p=p+C,因为x=1时,p=1所以C=0x/y'=y'y'=±√x,y=±(2/3)x^(3/2)+C代入初值y=(2/3)x^(3/2)+1/3或y=-(2/3)x^(3/2)+5/32.当p=...