Question

求一道高数题12.

Answer 1

该微分方程属于缺 x 型，即缺自变量型。
设 y' = p 则 y'' = dp/dx = (dp/dy)(dy/dx) = pdp/dy
微分方程化为 pdp/dy = 1+p^2
2pdp/(1+p^2) = 2dy, ln(1+p^2) = 2y+lnC1
1+p^2 = C1e^(2y)
p = ±√[C1e^(2y)-1] = dy/dx
dy/√[C1e^(2y)-1] = ±dx
记 √[C1e^(2y)-1] = u， 则 C1e^(2y) = 1+u^2
y = (1/2)ln[(1+u^2)/C1] = (1/2)[ln(1+u^2)-lnC1],
dy = (1/2)[2udu/(1+u^2)] = udu/(1+u^2)
dy/√[C1e^(2y)-1] = ±dx 化为 du/(1+u^2) = ±dx
arctanu = ±x + C2, u = tan(C2±x) = √[C1e^(2y)-1]

Answer 2

Answer 3

求微分方程 y''=1+(y')²的通解
解：令y'=dy/dx=p，则y''=dp/dx; 代入原式得：dp/dx=1+p²；
分离变量得：dp/(1+p²)=dx；积分之得：arctanp=x+c₁；
故有p=dy/dx=tan(x+c₁)；
∴通解y=∫tan(x+c₁)dx=∫tan(x+c₁)d(x+c₁)=-lncos(x+c₁)+c₂;