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### Topics - Hongling Liu

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##### 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)

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##### 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

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##### 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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