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Messages - Dang Tongbo

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Term Test 1 / Re: Problem 4 (afternoon)
« on: October 24, 2019, 01:38:59 PM »
(a) r^2 + 2r + 17 = 0
     r1 = -1 + 4i
     r2 = -1 - 4i
    so, y = c1*(e^(-x))*cos(4x) + c2*(e^(-x))*sin(4x))
    yp1 = A*(e^x)
    y'p1 = A*(e^x)
    y''p1 = A*(e^x)
    A*(e^x) + 2*A*(e^x) + 17*A*(e^x) = 40*A*(e^x)
                                                     A = 2
so yp1 = 2*(e^x)
   yp2 = A*sin(4x) + B*cos(4x)
   y'p2 = 4*A*cos(4x) - 4*B*sin(4x)
   y''p2 = -16*A*sin(4x) -16*B*cos(4x)
   -16Asin(4x) -16Bcos(4x) + 8Acos(4x) -8Bsin(4x) + 17Asin(4x) + 17Bcos(4x) = 130sin(4x)
  In this way, we can solve that: A = 2 and B = -16
 yp2 = 2sin(4x) -16cos(4x)
y = c1*(e^(-x))cos(4x) + c2*(e^(-x))sin(4x) +2*(e^x) + 2sin(4x) - 16cos(4x)
(b) y' = -C1*(e^(-x))cos(4x) -4C1*(e^(-x))sin(4x) -C2*(e^(-x))sin(4x) + 4C2(e^(-x))cos(4x) + 2*(e^x)+ 8cos(4x) + 64sin(4x)
     When y(0) = 0, y'(0) = 0,
      so, c1 = 14 c2 = 1
       y = 14*(e^(-x))cos(4x) +(e^(-x))sin(4x) +2*(e^x) + 2sin(4x)- 16cos(4x)

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Quiz-3 / Tut 0301 - Quiz 3
« on: October 13, 2019, 03:07:39 PM »
2y'' + 4y' -4y = 0 , y(0) = 0, y'(0) = 1.
Solution:
 2(r^2)+r-4= 0
so, r1 = ((-1)+(33)^(1/2))/4
     r2 = ((-1)+(33)^(1/2))/4
     y = c1*e^(((-1)+(33)^(1/2))/4)+c2*e^(((-1)-(33)^(1/2))/4). 
Because y(0)=0, so, c1+c2 = 0.
y' = -((-1)-(33)^(1/2))/4)*c1*e^(((-1)+(33)^(1/2))/4)-((1+(33)^1/2))*c2*e^(((-1)-(33)^(1/2))/4)
Because y'(0) = 1, so 1 = -(1-(33)^(1/2)/4)*c1 - (1+(33)^(1/2))/4)*c2
c1 = -c2,
so, -(1-(33)^(1/2)/4)*(-c2) - (1+(33)^(1/2))/4)*c2=1
        c2 = -(4/(2*(33)^1/2))
       c1 = 4/(2*(33)^1/2)
so, y = 4/(2*(33)^1/2)*e^((-1-(33)^1/2))/4)*t - (4/2*(33)^(1/2))/4)*t


3
Quiz-2 / Tut 5103- Quiz 2
« on: October 06, 2019, 05:23:05 PM »
Question: (x+2)* sin(y) + x*cox(y)*(dy/dx) = 0, u(x,y) = x*e^(x)
Solution:M(x,y) = (x+2)*sin(y),N(x,y) = cos(y)
             My = (x+2)*cos(y), Nx= cos(y)
             Because My is not equal to Nx, the equation is not exact.
             In this way, (x+2)*(x*e^(x))*sin(y)+ x^2 *(e^x)*cos(y)*(dy/dx)=0
             so, Lx(x,y) = (x+2)x(e^x)sin(y)
                  Ly(x,y) = (x^2)(e^x)cos(y)
                  L(x,y) = int((x+2)x*(e^x)sin(y)dx = (x^2)^(e^x)sin(y)+h(y)
                  Ly(x,y)= (x^2)(e^x)cos(y)+h'(y)= (x^2)(e^x)cos(y)
              h'(y) = 0
           As a result, (x^2)*(e^x)sin(y)=C
                 

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