Messages

1

Term Test 1 / Re: Problem 2 (main)

« on: October 23, 2019, 01:44:09 PM »

For b)
I choose c = 1
= xcosx(tan x + 1)
= xcosxtanx + xcosx
= x cosx(six/cosx) + xcosx
= xsinx + xcosx

2

Term Test 1 / Re: Problem 4 (main)

« on: October 23, 2019, 01:36:50 PM »

a)
First, solve the homogeneous system:
                   y’’-6y’+10y = 0
                    r^2-6r+10 = 0
                         R = 3+/-i
                   r1=3 + i  r2=3 - i. 
And
             Yc = C1e^(3x)cosx + C2e^(3x)sinx.

Now solve non homogeneous part:
                   y’’-6 y’+10 y = 2e^3x
LetYp = Ae^3x
               Then Yp’=3Ae^3x.   Yp’’ = 9Ae^3x
Plug in to the equation we have
                9Ae^3x-18Ae^3x + 10Ae^3x = 2e^3x
                               A = 2
So, Yp = 2e^3x
Now, we solve Y’’ - 6y’ +10y = 39cosx
                      Let Yp = Bcosx + Csinx
                     Then Yp’=-Bsinx + Ccosx   Yp’’ = -Bcosx - Csinx

Solve the linear equation:
                             9B-6C = 39
                              6B +9C = 0
                             B = 3, C = -2
Yp = 3cosx - 2sinx
Combining all we have:
Y= c1e^3xcosx + c2e^3xsinx + 2e^3x + 3cosx -2sinx

b)
As y(0) = 0 and y’(0) = 0,
Get the final solution
                           C1 = -5
                              C2 = 11
y = -5e^3xcosx + 11e^3xsinx + 2e^3x + 3cosx -2sinx

3

Quiz-4 / TUT0702 Quiz4

« on: October 18, 2019, 02:27:42 PM »

y'' +4y' = 3sin2t, y(0) = 0, y'(0)= -1

Sol:
r^2+4r=0
r(r+4)=0
r1=0,r2=-4
y=c1+c2e^(-4t)

y'' +4y' = 3sin2t
set Yp=Acos2t+Bsin2t
Yp'=-2Asin2t+2Bcos2t
Yp''=-4Acos2t-4Bsin2t
Plug in:
A= -3/10
B= -3/20
Yp(t)=-3/10cos2t-3/20sin2t
Y(t) = c1+c2e^(-4t)-3/10cos2t-3/20sin2t
plug in the initial values:
C1=1/8
C2=7/40
Therefore, the solution of this initial value problem is:
Y(t) = 1/8+7/40e^(-4t)-3/10cos2t-3/20sin2t

4

Quiz-3 / TUT0702 Quiz3

« on: October 11, 2019, 01:51:34 PM »

Find the differential equation has the general solution of :
  y = C1e^(-t/2) + C2e^-2t

Solution:
       The general solution has the form y = C1e^r1t + C2e^r2t
       we have r1 = -1/2.  r2 = -2
       (r + 1/2)(r+2) = 0
       r^2 + 5/2r + 1 = 0
       The differential equation has the from of y'' + p(t)y' +q(t)y = g(t)
       Therefore,
       the differential equation which has the general solution of has the general solution of y = C1e^(-t/2) + C2e^-2t is:
       Y'' + 5/2Y' + Y = 0
 

5

Quiz-2 / Re: TUT0702 Quiz2

« on: October 04, 2019, 03:10:23 PM »

Hi Qihui! Same as you until My ?= Nx
I tried My-Nx/M, My-Nx/N and Nx-My/XM-YN, all wrong. Did he said we don't need to count it? Just to show that the equation not exact should be fine, right?
Thank you

6

Quiz-2 / TUT0702 Quiz2

« on: October 04, 2019, 02:18:30 PM »

                                                       1+(x/y-siny)y' = 0
               M = 1,      N = x/y-siny
               My = o,    Nx = 1/y
My != Nx, the equation not exact.
               R1 = (My-Nx)/M = -1/y
               u = e^∫R1 dy
               u = y
Multiply u to both sides,
                y + (x - Ysiny)y' = 0
Now, the equation exact.
               there exist Φ(x, y), st.Φx = M
                                        Φ = xy +h(y)
                                     Φy = x + h'(y)
                                     h'(y) = -y siny
                                     h(y) = y cosy - siny
So, we have         Φ(x,y) = xy +ycosy - Siny
Therefore, the solution of the differential equation is :
                             xy + y cosy - siny = C