 | y'-(2/(1-x^2))y=-(1+x)/(1-x^2)
⇔y'-(2/(1-x^2))y=1/(x-1)
y'-2/(1-x^2)y=0 の解を求める。
∫1/ydy=∫2/(1-x^2)dx=∫{(1+x)+(1-x)}/{(1+x)(1-x)}dx …(1)
(1)右辺=∫{(1+x)+(1-x)}/{(1+x)(1-x)}dx=∫{1/(1-x)+1/(1+x)}dx=log|(x+1)/(x-1)|+C
(1)左辺=logy+C
y=A(x+1)/(x-1)>0
A→u=u(t)と置き換え
y'-(2/(1-x^2))y=1/(x-1)…(2) の解を求める。
y=u(x+1)/(x-1)
y'=u'{(x+1)/(x-1)}+u{-2/(x-1)^2}
(2)に代入
u'{(x+1)/(x-1)}+u{-2/(x-1)^2}-(2/(1-x^2))u*(x+1)/(x-1)=1/(x-1)
→u'=1/(x^2-1)
→u=(1/2)log|(x-1)/(x+1)|+C
y=u(x+1)/(x-1)={log|(x-1)/(x+1)|+A}(x+1)/{2(x-1)}//
|
