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### Messages - Hongling Liu

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1
##### Term Test 1 / Re: Problem 4 (afternoon)
« on: October 23, 2019, 06:45:36 AM »
y’’ +2y’ + 17y = 40e^x + 130sin3x
Solution:
a):
r^2 + 2r +17 =0
r1 = -1 + 4i, r2 = -1 -4i
∴y(x) = C1•e^-x•cos4x + C2•e^-x•sin4x
y’’ +2y’ + 17y = 40e^x
Let Yp(x) = A•e^x
Y’ = A•e^x, Y’’ = A•e^x
20•A•e^x = 40•e^x ∴A = 2

Where did you learn this crap? V.I.

2
##### Term Test 1 / Re: Problem 3 (afternoon)
« on: October 23, 2019, 06:44:52 AM »
y’’ - 5y’ + 6y = 52cos2x
Solution:
a):
r^2 - 5r + 6 = 0
r1 = 3
r2 = 2
∴y(x) = C1•e^3x + C2•e^2x
∵ y’’ - 5y’ + 6y = 52cos2x
set Yp(x) = Acos2x + Bsin2x
Y’ = -2Asin2x + 2Bcos2x
Y’’ = -4Acos2x - 4Bsin2x
-4Acos2x - 4Bsin2x -5(-2Asin2x + 2Bcos2x) + 6(Acos2x + Bsin2x) = 52cos2x
A = 1
B = -5
∴Y(x) = C1•e^3x + C2•e^2x + cos2x - 5sin2x
b):
When Y(0) = 0, Y’(0)= 0
∴C1 =12, C2= -13
∴Y(x) = 12•e^3x - 13•e^2x + cos2x - 5sin2x

3
##### Term Test 1 / Re: Problem 2 (afternoon)
« on: October 23, 2019, 06:43:41 AM »
2:
(2x+1)xy’’ + (2x+2)y’ -2y = 0
Solution:
a):
y’’ + [(2x+2)/(2x+1)x]y’ -[2/(2x+1)x]y= 0
W = C•e^∫-[(2x+2)/(2x+1)x]dx
W = C•(2x+1/x^2) let C = 0
W  = (2x+1/x^2)
b):
let another solution is y2
(1+x)y’2-y2 = (2x+1/x^2)
y2 = -1/x + C•(x+1)
Let C=0
y2 = -1/x + (x+1)
c):
y(x) = C1(x+1) + C2[-1/x + (x+1)]
From the problem we can get
C1 = -2
C2 = 1
∴ y(x) = -2(x+1) + [-1/x + (x+1)]
∴ y(x) = -x -(1/x) - 1

4
##### Term Test 1 / Re: Problem 1 (afternoon)
« on: October 23, 2019, 06:42:50 AM »
-y^2•sinxy + (-xy•sinxy + 2cosxy + 3y)•y’ = 0
Solution:
a):
My = -2y•sinxy - x•y^2•cosxy
Nx = -3y•sinxy - x•y^2•cosxy
My≠Nx
∴R1 = (My - Nx)/M = -1/y
u = e^(∫-R1dy) = y
∴ -y^3•sinxy + y•(-xy•sinxy + 2cosxy + 3y)•y’ = 0
My = -3y^2•sinxy - x•y^3•cosxy
Nx = -3y^2•sinxy - x•y^3•cosxy
∴My =Nx
ψ(x,y) = ∫Mdx = y^2•cosxy + h(y)
ψy = N
ψy = 2y•cosxy - x•y^2sinxy + h’(y)
∴h’(y) = 3y^2
h(y) = y^3
ψ(x,y) = y^2•cosxy + y^3 = C
b):
y(π/3) = 1 ∴C= 3/2
∴ ψ(x,y) = y^2•cosxy + y^3 = 3/2

Correct but looks really ugly . V.I.

5
##### 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)

6
##### 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

7
##### 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

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