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### Messages - Zhangxinbei

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##### 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-1 / Re: TUT0702 quiz and solutions
« on: October 04, 2019, 03:20:52 PM »
Sure!
y'+ 2/t y = sint/t
y' + p(t)y = g(t)
p(t) = 2/t.  G(t) = sint/t
u = e^∫2/t dt = t^2
multiply t^2 to both sides:
t^2y' + 2ty = t sint
(t^2y)' = t sint
t^2y = ∫t sint dt
y = (-t cost + pint +c)/t^2

SO, given t>0, y->o, as t -> infinity

6
##### 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

7
##### 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

8
##### Quiz-1 / TUT0702 quiz and solutions
« on: September 27, 2019, 04:39:52 PM »
Find the initial value problem.  Y' = 2x/1+y^2, y(2)=0.
Using separable:
dy/dx = (2x)/(1+y^2)
integral on both sides:
∫ 1+y^2 dy=∫ 2x dx
y+(y^3)/3 = x^2 + c
as y(2) = 0, plug into the function:
0 = 4+c
c = -4
Therefore,  y+(y^3)/3 = x^2 -4

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