Question

如图，这个微分方程怎么求，详细过程谢谢

Answer 1

解:微分方程为(1-u²)f'(u)+2f(u)=u，化为

f'(u)+2f(u)/(1-u²)=u/(1-u²)，两边同时 乘以(1+u)/(1-u)，有

f'(u)(1+u)/(1-u)+2f(u)/(1-u)²=u/(1-u)²，

[f(u)(1+u)/(1-u)]'=1/(u-1)²+1/(u-1)，

f(u)(1+u)/(1-u)=ln|u-1|-1/(u-1)+c

(c为任意常数)，方程的通解为

f(u)=(1-u)ln|u-1|/(1+u)+1/(u+1)+

c(1-u)/(1+u)

Answer 2

p(u)=2/(1-u^2)
∫p(u)du
=∫2/(1-u^2) du
=∫[1/(1-u) +1/(1+u)] du
=ln|(1+u)/(1-u)| +C
e^[∫p(u)du] =(1+u)/(1-u)
//
(1-u^2)f'(u) +2f(u) = u
f'(u) +[2/(1-u^2) ]f(u) = u/(1-u^2)

两边乘以(1+u)/(1-u)
[(1+u)/(1-u)] .[f'(u) +[2/(1-u^2) ]f(u)] =[(1+u)/(1-u)].[ u/(1-u^2)]
d/du { (1+u)/(1-u)] .f(u) } =u/(1-u)^2
[(1+u)/(1-u)] .f(u) = ∫u/(1-u)^2 du
=-∫u d[1/(1-u)]
=-u/(1-u) +∫[1/(1-u)] du
=-u/(1-u) -ln|1-u| +C
f(u) =[(1-u)/(1+u)] . [-u/(1-u) -ln|1-u| +C]

Answer 3

你化成如下形式可以直接用公式