第1个回答 2022-03-19
设 dy/dx = y' = p , 则 dp/dx = (dp/dy)(dy/dx) = pdp/dy,
微分方程 y′′y^3+1 = 0 化为 y^3pdp/dy = -1, 2pdp = -2dy/y^3,
p^2 = y^(-2) + C1, x = 1 时 , y = 1, y' = p = 0 代入得 C1 = -1，
p^2 = y^(-2) - 1 ， p = dy/dx = ±√[y^(-2) - 1] = ±√(1-y^2)/y
±ydy/√(1-y^2) = dx, ±(1/2)d(1-y^2)/√(1-y^2) = dx,
±√(1-y^2) = x + C2, x = 1 时 , y = 1, 代入得 C2 = -1,
特解 ±√(1-y^2) = x -1
微分方程 y′′y^3+1 = 0 化为 y^3pdp/dy = -1, 2pdp = -2dy/y^3,
p^2 = y^(-2) + C1, x = 1 时 , y = 1, y' = p = 0 代入得 C1 = -1，
p^2 = y^(-2) - 1 ， p = dy/dx = ±√[y^(-2) - 1] = ±√(1-y^2)/y
±ydy/√(1-y^2) = dx, ±(1/2)d(1-y^2)/√(1-y^2) = dx,
±√(1-y^2) = x + C2, x = 1 时 , y = 1, 代入得 C2 = -1,
特解 ±√(1-y^2) = x -1
第2个回答 2023-03-19
解:微分方程为y"y³+1=0，化为y"=-1/y³，
2y'y"=-2y'/y³，y'²=1/y²+c(c为任意常数)
∵y(1)=1，y'(1)=0 ∴有0=1/1²+c，得:c=-1
∴微分方程为y'²=1/y²-1，-yy'/√(1-y²)=±1，√(1-y²)=±x±a，1-y²=(x+a)²(a为任意常数)，微分方程的通解为(x+a)²+y²=1
∴有(1+a)²+1²=1，得:a=-1 ∴微分方程的特解为(x-1)²+y²=1
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