Question

高数微分方程求特解，谢谢

如下图

Answer 1

p=y', y'' = dy'/dx = dy'/dy dy/dx = pdp/dy
pdp/dy +p^2 =1
pdp/(1-p^2) = dy
dp [1/(1-p) - 1/(1+p)] = 2dy
ln (1-p) - ln(1+p) = 2y +C

y=0时，p=0代人得到
C=0
ln (1-p) - ln(1+p) = 2y
(1-p)/(1+p) = e^(2y)
2/(1+p) = 1+e^(2y)
p=2/(1+e^(2y)) -1
y'=2/(1+e^(2y)) -1 = [1-e^(2y)]/(1+e^(2y))

dy [1+e^(2y)]/(1-e^(2y)) = dx
令t=e^y, 上式变为
(1+t^2)/t(1-t^2) dt =dx
令(1+t^2)/t(1-t^2) = a + b/t + c/(1+t) + d/(1-t)可以求出abcd
然后就可以求出左边的积分，得到特解