Toronto Math Forum

MAT244--2019F => MAT244--Test & Quizzes => Quiz-1 => Topic started by: Xuefeng Fan on September 27, 2019, 02:00:07 PM

Title: TUT 0401
Post by: Xuefeng Fan on September 27, 2019, 02:00:07 PM
Find the general solution of the given equation:
xy' = (1-y^2)^1/2
solution: Because Separable,
therefore x(dy/dx) = (1-y^2)^1/2,
Rearrange: \int(1/(1-y^2)^1/2)*dy =\int(1/x)dx, where x not equal 0, y does not equal to positive or negative 1,
therefore: arcsin(y) = ln|x| + C,
therefore general solution: y=sin(ln|x| + C) where x not equal to 0, y not equal to positive 1 or negative 1.