Question

求下列微分方程通解

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 + C1
x = 0 时 , y = 0, y' = p = 0, 得 C1 = 0
1-p^2 = e^(-2y)
p = dy/dx = ±√[1-e^(-2y)]
dy/√[1-e^(-2y)] = ±dx. e^ydy/√[e^(2y)-1] = ±dx
令 e^y = secu, 则 secutanudu/tanu = ±dx
secudu = ±dx, ln|secu+tanu| = ±x + C2
ln{e^y+√[e^(2y)-1]} = ±x + C2
x = 0 时, y = 0, 得 C2 = 0
特解是 e^y+√[e^(2y)-1] = e^(±x)

Answer 2

设p=y',
y"=dp/dx=dp/dy*dy/dx=pdp/dy
pdp/dy+p²=1
∫p/(1-p²)dp=y
-1/2*ln|p²-1|=y+C
x=0,y=0,p=y'(0)=0
ln|p²-1|=-2y
p²-1=e^(-2y)
p=√[1+e^(-2y)]
∫e^y/√[(e^y)²+1]dy=x
ln[e^y+√(1+e^(2y))]=x+lnC1
x=0,y=0,C1=√2+1
e^y+√[1+e^(2y)]=(√2+1)e^x
即为所求特解

Answer 3

这题可以用带入法，直接将题一上的式子带入二式，在拆开分解