(1-x)dy=(y-1)dx- уравнение с разделяющимися переменными
dy/(y-1)=dx/(1-x)
∫ dy/(y-1)= -∫ dx/(x-1)
ln|y-1|=-ln|x-1|+lnC
ln|y-1|=lnC/|(x-1)|
y-1=C/(x-1)
[b](2;3)[/b]
3-1=C/(2-1)
x=2
y-1=2/(x-1)
[b]y=2/(x-1) +1[/b]
2)
y`=y-4
dy/dx=(y+4)
dy/(y+4)=dx
∫ dy/(y+4)= ∫ dx
[b]ln|y+4|=x+C[/b]
3)
y`=1/(2y)
dy/dx=(1/2y)
2ydy=dx
∫ 2ydy= ∫ dx
y^2=x+C
[b](2;1)[/b]
1=2^2+C
C=-3
[b]y^2=x-3
[/b]
4)
y`=dy/dx
(dy/dx)*sqrt(1-x^2)=x
dy=xdx/sqrt(1-x^2)
∫ dy= ∫ xdx/sqrt(1-x^2)
∫ dy= (-1/2)∫(-2 xdx)/sqrt(1-x^2)
∫ dy= (-1/2)∫(1-x^2)^(-1/2)d(1-x^2) табл интеграл ∫ u^(-1/2)du=2sqrt(u)
y=-sqrt(1-x^2) +C
[b](0;1)[/b]
1=sqrt(1-0)+C
C=0
[b]y=-sqrt(1-x^2)
[/b]