Author Topic: tut0501 quiz2  (Read 303 times)

Hongling Liu

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tut0501 quiz2
« on: October 04, 2019, 02:00:06 PM »

      (x+2)sin(y) + x⋅cos(y)y’= 0. μ =x⋅e^x
     Solution:
     My =(x+2)cos(y)
     Nx = cos(y)
     ∴My ≠ Nx it is not Exact
     Multiply μ
      x⋅e^x (x+2)sin(y) + x⋅e^x⋅x⋅cos(y)y’= 0
     My = x(x+e)e^x⋅cos(y)
     Nx = x(x+e)e^x⋅cos(y)
     Now My = Nx and it is Exact
     ψy = N  ψ = ∫ N ⅾy =∫ x⋅e^x⋅x⋅cos(y) ⅾy
     ∴ψ = x^2⋅e^x⋅sin(y) + h(x)
     ∵ψx = M   ψx = (2+x)⋅x⋅e^x⋅sin(y) +h’(x) = M
     ∴h’(x) =0 ∴h(x) = C
     ψ(x,y) = x^2⋅e^x⋅sin(y) = C