Topics

1

Quiz-4 / 0501 quiz4

« on: October 18, 2019, 02:00:51 PM »

   t^2y’’ + 3ty’ + 1.25y = 0
   Solution:
   lnx =  t
   d^2y/dx^2 + (3-1)dy/dx +5/4y = 0
   r^2 + 2r + 5/4 =0
   r1 = -2 + i     t2 = -2 - i
   y(x) = C1⋅e^(-2)⋅cos(x) + C2⋅e^(-2)⋅sin(x)
   ∴y(t) = C1⋅(1/2)⋅cos(lnt) + C2⋅(1/2)⋅sin(lnt)

2

Quiz-3 / quiz3 0501

« on: October 11, 2019, 02:02:10 PM »

    y‘’ - 2y’ - 2y = 0
    Solution:
    r^2 - 2r - 2= 0
    r1 = 1 +√3
    r2 = 1- √3
    y = C1⋅e^(1+√3)⋅t + C2⋅e^(1-√3)⋅t

3

Quiz-2 / 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