y”=(y’/x)/(1+ln(y'/x)) y'(1)=1 y(1)=1/2
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y”=(y’/x)/(1+ln(y'/x)) y'(1)=1 y(1)=1/2
p=y',p'(x+p)=p
(dp/dx)(x+p)=p
1.p≠0时 (dx/dp)=x/p+p
令z=x/p,x=zp,x'=z+z'p
z+z'p=z+p
z'p=p
x/p=p+C,因为x=1时,p=1所以C=0
x/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)+5/3
2.当p=0时,无法满足y'(1)=1的条件
故结果为:
y=(2/3)x^(3/2)+1/3或y=-(2/3)x^(3/2)+5/3
(dp/dx)(x+p)=p
1.p≠0时 (dx/dp)=x/p+p
令z=x/p,x=zp,x'=z+z'p
z+z'p=z+p
z'p=p
x/p=p+C,因为x=1时,p=1所以C=0
x/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)+5/3
2.当p=0时,无法满足y'(1)=1的条件
故结果为:
y=(2/3)x^(3/2)+1/3或y=-(2/3)x^(3/2)+5/3
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