X+Y–2+(1–X)Y'=0
y`+(1/(1–x))·y=(2–x)/(1–x)
y=u·v
y`=u`·v+u·v`
u`·v+u·v`+(1/(1–x))·u·v=(2–x)/(1–x)
u`·v+u·(v`+(1/(1–x))·v)=(2–x)/(1–x)
{v`+(1/(1–x))·v=0⇒ dv/v=dx/(x–1) ⇒ ∫ dv/v= ∫ dx/(x–1) ⇒ln|v|=ln|x–1|
{u`·v=(2–x)/(1–x) ⇒u`·(x–1)=(2–x)/(1–x)
⇒ u= ∫ (x–2)dx/(x–1)2= ∫ (x–1)dx/(x–1)2– ∫dx/(x–1)2=
= ∫ (dx/(x–1)– ∫dx/(x–1)2=ln|x–1|+ 1/(x–1) + C
y=u·v=(ln(x–1)+ 1/(x–1) + C)·(x–1)=(x–1)ln(x–1)+C·(x–1) + 1 – о т в е т.