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### Messages - Yiran Zhu

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##### Quiz-3 / Re: Q3 TUT 5101
« on: October 12, 2018, 06:21:57 PM »
This is the hand written solution. I am still trying to figure out how to convert latex into plain text.

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##### Quiz-2 / Re: Q2 TUT 0301, TUT 5101 AND 5102
« on: October 05, 2018, 05:43:18 PM »
Since the equation is not exact, we need to find a function µ(x, y), such that

µ(3x2y + 2xy + y3) + µ(x2 + y2)y’ = 0 is exact.

M = µ(3x2y + 2xy + y3)      N = µ(x2 + y2)

Then  My  = µy’(3x2y + 2xy + y3) + µ(3x2 + 2x + 3y2)
Nx = µx’(x2 + y2) + µ(2x)

Let My = Nx , we get µy’(3x2y + 2xy + y3) + µ(3x2 + 3y2) = µx’(x2 + y2)

First suppose µ is a function of x only.

Then µy’ = 0, we get
µ(3x2 + 3y2) = µx’(x2 + y2)
3µ = ∂µ/∂x

By separable equation,
∫(1/µ) ∂µ = ∫3 ∂x
ln(µ) = 3x
µ = e3x (the integration factor)

By theorem, there exist ϕ(x,y) , such that , ϕx = M, ϕy = N

ϕ = ∫e3x(3x2y + 2xy + y3) = e3xx2y + (e3xy3)/3 + h(y)
ϕy = e3xx2 + h'(y) = e3xx2 + e3xy2
Thus, h'(y) = e3xy2, h(y) = (e3xy3)/3 + C

Therefore, the general solution is ϕ = e3xx2y + (e3xy3)/3 = C

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##### Quiz-2 / Re: Q2 TUT 0101, TUT 0501 and TUT 0801
« on: October 05, 2018, 05:41:00 PM »
Sorry, I posted to the wrong quiz question

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