x^2 dx + y(x-1) dy = 0
The answer will be
(x+1)^2 + y^2 + 2 ln |c(x-1)|=0
Thank you!
The answer will be
(x+1)^2 + y^2 + 2 ln |c(x-1)|=0
Thank you!
-
[x^2/(x-1)]dx + ydy = 0
[(x+1)+1/(x-1)]dx + ydy=0
integrate both sides
(x+1)^2/2+ ln|x-1| + y^2/2 = A , A = constant
or
(x+1)^2 + y^2 + 2 ln |c(x-1)|=0 , c = constant
[(x+1)+1/(x-1)]dx + ydy=0
integrate both sides
(x+1)^2/2+ ln|x-1| + y^2/2 = A , A = constant
or
(x+1)^2 + y^2 + 2 ln |c(x-1)|=0 , c = constant