Question

考研数学，高数，微分方程，问题如图

Answer 1

x = 0 时，微分方程无解。
x ≠ 0 时，记 p = dy/dx, 则微分方程变为
xdp/dx = -(1/2) √(1+p^2),
dp/√(1+p^2) = -(1/2)dx/x
令 p = tanu，则
secudu = -(1/2)dx/x
初始条件是 x < 0 ， 故求 x < 0 时的特解。
ln(secu+tanu) = (1/2)ln(-x) + lnC
secu + tanu = C√(-x)
p+√(1+p^2) = C√(-x)
p(-1) = 1, 得 C = 1+√2
p+√(1+p^2) = (1+√2)√(-x)
1+p^2 = [ (1+√2)√(-x)-p]^2 = -(3+2√2)x -2p (1+√2)√(-x)+p^2
2p(1+√2)√(-x) = -1-(3+2√2)x
p = dy/dx = -[1+(3+2√2)x]/[2(1+√2)√(-x)]
dy = [-(√2-1)/2] [1/√(-x)-(3+2√2)√(-x)]dx
y = [-(√2-1)/2] [-2√(-x)+(2/3)(3+2√2)(-x)^(3/2)] + C1
y(-1) = 0, 得 C1 = [-(√2-1)/2] (4√2/3) = 2(2-√2)/3
得 y = [(√2-1)/2] [2√(-x)-(2/3)(3+2√2)(-x)^(3/2)] + 2(2-√2)/3