一道高数题

Question

求y''(x+y'^2)=y'满足初始条件y(1)=y'(1)=1的特解怎么做

Answer 1

令y'=p，则y''=p'
p'(x+p²)=p
dp/dx = p/(x+p²)
xdp + p²dp = pdx
dp = (pdx-xdp)/p² = d(x/p)
则p = x/p + C
解出p = C±√(C²+x)
即y' = C±√(C²+x)
y'(1)=1，则1 = C±√(C²+1)，解出C=0
即y' = √x
积分得y = 2/3 √x³ + C'
y(1)=1，则1 = 2/3 + C'，C' = 1/3
特解是y = 2/3 √x³ + 1/3

Answer 2

不懂诶
暂时初中滴~
不会
几年后再告诉你~！
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