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Messages - Zhiya Lou

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1
Final Exam / Re: FE-P2
« on: December 14, 2018, 09:28:44 AM »
I think Cui calculated the homogeneous solution wrong:
$(r-1)(r-1)(r-1) = r^3 -3r^2+3r+1$

Actually:
$r^3 -3r^2+4r-2= (r-1)(r^2-2r-2)$ = 0
$r=1$ or $r=1-i, 1+i$

So, Homogeneous solution is
$y= c_1e^t + c_2e^t\cos(t) + c_3e^t\sin(t)$

Non homogenous part for $10e^t$
We should assume $Y= Ate^t$
$Y’=Ate^t+Ae^t$
$Y’’=Ate^t+2Ae^t$
$Y’’’=Ate^t+3Ae^t$

$Y’’’-3Y’’+4Y’-2Y =(A-3A+4A-2A)te^t+(3A-6A+4A-2A)e^t =-Ae^t =10e^t$
$A =-10, Y=-10te^t$

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Quiz-7 / Re: Q7 TUT 0601
« on: November 30, 2018, 04:30:46 PM »
Solution

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Quiz-6 / Re: Q6 TUT 0601
« on: November 20, 2018, 08:55:58 AM »
Rotation direction for complex roots: look at original matrix, since b= -5 <0 and c=1>0, so it is counterclockwise.
When the real part is positive, spiral outward, unstable; when the real part is negative, spiral inward, stable.

when $\alpha$ = $\sqrt{20}$  (it's positive, outward, unstable)
It is in repeated root case, but only has one independent eigenvector, therefore, we could graph in the direction of this eigenvector, following the direction of counterclockwise as the same for complex roots.
Similar for $\alpha$ = $-\sqrt{20}$, it's negative, inward, stable, counterclockwise still.

4
Quiz-6 / Re: Q6 TUT 0601
« on: November 17, 2018, 04:08:44 PM »
here is my solution

5
Quiz-5 / Re: Q5 TUT 0801
« on: November 02, 2018, 04:13:29 PM »
Let $x_1= u, x_2=u'$
Then substitute it into original equation:
$x_2'+0.25x_2+4x_1 = 2\cos(3t)$

So, we can transform into the system:
$x_1'=x_2$
$x_2'+0.25x_2+4x_1 = 2\cos(3t)$

With given initial value:$x_1(0)=1, x_2(0)= -2$

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Quiz-5 / Re: Q5 TUT 0601
« on: November 02, 2018, 03:57:41 PM »
Not sure.... but TUT 0601 actually have this question:

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Quiz-4 / Re: Q4 TUT 0601
« on: October 26, 2018, 09:55:34 PM »
First, solve the homogeneous solution:
$r^2+1=0$
$r_1=i, r_2= -i$
$y_1= cos(t); y_2 = sin(t)$
So, $W=y_1y_2' - y_2y_1'= cos^2(t) + sin^2(t) = 1$

Second, solve for particular solution:
$Y=u_1y_1 + u_2y_2$
$u_1(t) = -\int sin(t)tan(t) dt\ = -\int sec(t)(1-cos^2(t))dt = -\int sec(t) - cos(t) dt = -ln(tan(t) + sec(t)) +sin(t)$

$u_2(t) = \int y_1(t) tan(t) dt = \int cos(t)tan(t) dt = \int sin(t) dt = -cos(t)$

Therefore, $Y= [-ln(tan(t) + sec(t)) +sin(t)](cos(t)) + (-cos(t)) (sin(t))$

General Solution:
$y(t)= c_1cos(t) + c_2sin(t) - ln(tan(t)+sec(t))(cos(t))$

8
Term Test 1 / Re: TT1 Problem 1 (afternoon)
« on: October 16, 2018, 09:50:37 AM »
$M_y= -2yx-2x^2$
$N_x= -4xy -3x^2$

M$_y$ $\neq$ N$_x$ so the original equation is not exact

Let $\mu$ only depends on x

$\mu'(x)$ = $\frac{M_y - N_x}{N}$ $\mu(x)$

$\mu'(x)$  = $\frac{-2xy -2x^2 + 4xy + 3x^2 }{-x(2xy + x^2)}$  $\mu(x)$

$\mu'(x)$  = $\frac{-1}{x}$ $\mu(x)$

$\mu(x)$ = $\frac{1}{x}$

multiply $\mu(x)$ to both side:

($\frac{1}{x}$ - $y^2$ -2xy) -(2xy + $x^2$)$y'$ = 0    (Now M$_y$ = N$_x$, equation is exact)

There exist $\psi(x,y)$  st

$$\psi(x, y) = \int{\frac{1}{x} - y^2 - 2xy} \mbox{d}x = \ln x - xy^2 -x^2y + h(y)$$
$$\frac{\partial \psi}{\partial y} = -2yx - x^2 + h'(x) = -2xy - x^2$$
$$\mbox{Therefore, } h'(x) \Longrightarrow h(x) = c$$

$$\ln x - xy^2 - x^2y = c$$Given $$y(1) = 1$$ when $$x = 1, y = 1$$ such that $\ln 1 - 1 - 1 = c, \therefore c = -2$ $$\ln x - xy^2 - x^2 y = -2$$

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Term Test 1 / Re: TT1 Problem 3 (main)
« on: October 16, 2018, 09:06:33 AM »
scan solution
(sorry, almost post the same time, I didn't notice the perfect solution is posted before me)

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