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### Topics - Zhangxinbei

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##### 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

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##### 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

3
##### 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

4
##### 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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