Toronto Math Forum

MAT244--2019F => MAT244--Test & Quizzes => Quiz-1 => Topic started by: AllanLi on September 27, 2019, 02:01:31 PM

Title: quiz 1 tut 0401
Post by: AllanLi on September 27, 2019, 02:01:31 PM
       xy'=(1-y^2)^(1/2)
 write y' into dy/dx then we get
  x * (dy/dx) = (1-y^2)^(1/2)
 Then we rearrange the equation, let x,dx and y,dy on the same side.
  we get 1/x * dx = (1-y^2)^(-1/2) * dy
 by taking integral on both side,
    ln|x|+C = arcsin(y)
therefore y = sin(ln|x|+C)
Title: Re: quiz 1 tut 0401
Post by: AllanLi on September 27, 2019, 05:05:09 PM