Question

求微分方程y''＋y'^2＋1=0的通解。 答案给的是设y'为p,则y''为p' 但是y''不

Answer 1

设 y' = dy/dx = p(x), 则 y'' = d^2y/dx^2 = dp(x)/dx
微分方程 y''＋y'^2＋1 = 0 化为 dp/dx = 1-p^2
dp/(1-p^2) = dx, [1/(1-p) + 1/(1+p)]dp = 2dx
ln[(1+p)/(1-p)] = 2x + lnC, (1+p)/(1-p) = Ce^(2x)
p = dy/dx = [Ce^(2x)-1]/[Ce^(2x)+1]
y = ∫[Ce^(2x)-1]/[Ce^(2x)+1]dx
令 e^(2x) = u， 则 x = (1/2)lnu, dx = du/(2u)
y = (1/2)∫[Cu-1]/[u(Cu+1)]du = (1/2)∫[2C/(Cu+1) - 1/u]du
= (1/2)[2ln(Cu+1) - lnu] + C1 = ln[Ce^(2x)+1] - x + C1

Answer 2

设y'=p
y''=dp/dx=dp/dy*dy/dx=pdp/dy
原式变为
pdp/dy+p^2+1=0
pdp/(p^2+1)=-dy
1/2ln(p^2+1)=-y+C1
p^2+1=C2e^(-2y)
p^2=C2e^(-2y)-1
p=±√[C2e^(-2y)-1]=dy/dx
只能这样子了