Author Topic: quiz 1 tut 0401  (Read 824 times)

AllanLi

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quiz 1 tut 0401
« 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)

AllanLi

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Re: quiz 1 tut 0401
« Reply #1 on: September 27, 2019, 05:05:09 PM »