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Home Assignment 2 / Re: Assignment 2.2 Problem3 (16)
« on: January 24, 2019, 10:27:46 AM »
Could you just post the solution? Could really learn from examples instead of words.
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MAT244--Misc / When should we expect the marks?
« on: December 20, 2018, 12:42:51 PM »
When should be expect the marks for final, bonuses and quiz 5 update? Thanks.
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Final Exam / Re: FE-P5
« on: December 14, 2018, 10:37:07 AM »
$$J=\begin{bmatrix}
6x+2y-30 & 2x\\
-y & 4y-x-6
\end{bmatrix}$$
at (0,0):
$$J=\begin{bmatrix}
-30 & 0\\
0 & -6
\end{bmatrix}$$
diagonal matrix with negative eigenvalues => stable node
at (0,3):
$$J=\begin{bmatrix}
-24 & 0\\
-3 & 6
\end{bmatrix}$$
triangular matrix with eigenvalues -24 and 6 => saddle
at (10,0):
$$J=\begin{bmatrix}
30 & 20\\
0 & -16
\end{bmatrix}$$
diagonal matrix with eigenvalues 30 and -16 => saddle
at (6,6):
$$J=\begin{bmatrix}
18 & 12\\
-6 & 12
\end{bmatrix}$$
eigenvalues are $15+3i\sqrt{7}, 15-3i\sqrt{7}$ => unstable spiral
6x+2y-30 & 2x\\
-y & 4y-x-6
\end{bmatrix}$$
at (0,0):
$$J=\begin{bmatrix}
-30 & 0\\
0 & -6
\end{bmatrix}$$
diagonal matrix with negative eigenvalues => stable node
at (0,3):
$$J=\begin{bmatrix}
-24 & 0\\
-3 & 6
\end{bmatrix}$$
triangular matrix with eigenvalues -24 and 6 => saddle
at (10,0):
$$J=\begin{bmatrix}
30 & 20\\
0 & -16
\end{bmatrix}$$
diagonal matrix with eigenvalues 30 and -16 => saddle
at (6,6):
$$J=\begin{bmatrix}
18 & 12\\
-6 & 12
\end{bmatrix}$$
eigenvalues are $15+3i\sqrt{7}, 15-3i\sqrt{7}$ => unstable spiral
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MAT244--Misc / Re: Quiz 5 explanation (was Unfair quiz)
« on: November 30, 2018, 04:46:03 PM »
Can we have a followup for this?
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Quiz-7 / Re: Q7 TUT 0601
« on: November 30, 2018, 04:39:09 PM »
Solve $1-y = 0\Rightarrow y=1$ and $x^2 - y^2 = (x+y)(x-y) = 0$, substituting $y=1$, we have 2 critical points $(-1, 1), (1,1)$
Computing the Jacobian yields
\begin{align}
J =
\begin{bmatrix}
F_x & F_y\\
G_x & G_y
\end{bmatrix}
=
\begin{bmatrix}
0 & -1\\
2x & -2y
\end{bmatrix}
\end{align}
Plugging in both critical points we have 2 linear systems, first
\begin{align}
x' =
\begin{bmatrix}
0 & -1\\
-2 & -2
\end{bmatrix}
x
\end{align}
solving for eigenvalues yields $\lambda_1 = -1 + \sqrt{3}\, \lambda_2 = -1 - \sqrt{3}$, since $\sqrt{3} > 1$, we have $\lambda_1 > 0 > \lambda_2$, so we conclude locally $(-1, 1)$ is a saddle.
Plug in $(1,1)$
\begin{align}
x' =
\begin{bmatrix}
0 & -1\\
2 & -2
\end{bmatrix}
x
\end{align}
solving for eigenvalues yields $\lambda_1 = -1 + i\, \lambda_2 = -1 - i$, a complex conjugate with negative real parts, we conclude locally $(1, 1)$ is an asymptotically stable spiral.
Computing the Jacobian yields
\begin{align}
J =
\begin{bmatrix}
F_x & F_y\\
G_x & G_y
\end{bmatrix}
=
\begin{bmatrix}
0 & -1\\
2x & -2y
\end{bmatrix}
\end{align}
Plugging in both critical points we have 2 linear systems, first
\begin{align}
x' =
\begin{bmatrix}
0 & -1\\
-2 & -2
\end{bmatrix}
x
\end{align}
solving for eigenvalues yields $\lambda_1 = -1 + \sqrt{3}\, \lambda_2 = -1 - \sqrt{3}$, since $\sqrt{3} > 1$, we have $\lambda_1 > 0 > \lambda_2$, so we conclude locally $(-1, 1)$ is a saddle.
Plug in $(1,1)$
\begin{align}
x' =
\begin{bmatrix}
0 & -1\\
2 & -2
\end{bmatrix}
x
\end{align}
solving for eigenvalues yields $\lambda_1 = -1 + i\, \lambda_2 = -1 - i$, a complex conjugate with negative real parts, we conclude locally $(1, 1)$ is an asymptotically stable spiral.
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Quiz-5 / Re: Q5 TUT 0601
« on: November 03, 2018, 10:54:19 AM »
This is in fact not the question TUT0601 got, we had the question from Zhiya's post. I think it's meant to be for next week, since the TA just did it in tutorial and nobody had practiced it.
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Quiz-4 / Re: Q4 TUT 5102
« on: October 26, 2018, 08:35:54 PM »
To verify take $y_1 = e^t, y_2 = t$, then
\begin{align*}
y_1' = e^t,\ &y_2' = 1\\
y_1'' = e^t,\ &y_2'' = 0
\end{align*}
Plug in to the original equation, we check both are 0
\begin{align*}
(1-t)e^t + te^t - e^t &= 0\\
t - t &= 0
\end{align*}
Now take the Wronskian
\begin{align*}
W = e^t - te^t = (1-t)e^t
\end{align*}
Use the variation of parameters equation to find
\begin{align*}
u_1 &= -\int\frac{y_2g}{W}\\
&= -2\int te^{-2t}dt\\
&= (t+\frac{1}{2})e^{-2t}
\end{align*}
\begin{align*}
u_2 &= \int\frac{y_1g}{W}\\
&= 2\int e^{-t}dt\\
&= -2e^{-t}
\end{align*}
Putting the pieces together, we have the particular solution
\begin{align*}
Y &= (t+\frac{1}{2})e^{-t} - 2te^{-t}\\
&= \frac{1}{2}e^{-t} - te^{-t}\\
&= e^{-t}(\frac{1}{2} - t) \\
&= -\frac{1}{2}(2t-1)e^{-t}\\
\end{align*}
\begin{align*}
y_1' = e^t,\ &y_2' = 1\\
y_1'' = e^t,\ &y_2'' = 0
\end{align*}
Plug in to the original equation, we check both are 0
\begin{align*}
(1-t)e^t + te^t - e^t &= 0\\
t - t &= 0
\end{align*}
Now take the Wronskian
\begin{align*}
W = e^t - te^t = (1-t)e^t
\end{align*}
Use the variation of parameters equation to find
\begin{align*}
u_1 &= -\int\frac{y_2g}{W}\\
&= -2\int te^{-2t}dt\\
&= (t+\frac{1}{2})e^{-2t}
\end{align*}
\begin{align*}
u_2 &= \int\frac{y_1g}{W}\\
&= 2\int e^{-t}dt\\
&= -2e^{-t}
\end{align*}
Putting the pieces together, we have the particular solution
\begin{align*}
Y &= (t+\frac{1}{2})e^{-t} - 2te^{-t}\\
&= \frac{1}{2}e^{-t} - te^{-t}\\
&= e^{-t}(\frac{1}{2} - t) \\
&= -\frac{1}{2}(2t-1)e^{-t}\\
\end{align*}
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MAT244--Misc / Test 1 marks?
« on: October 25, 2018, 12:19:40 PM »
When are we getting the marks back?
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Term Test 1 / Re: TT1 Problem 4 (main)
« on: October 16, 2018, 09:14:51 AM »
Let me retype it in MathJax
First solve homogenous complement equation
\begin{align*}
y'' + 2y' + 5y &= 0\\
r^2 + 2r + 5 &= 0\\
r_1 = 1+2i,\ &r_2 = 1-2i
\end{align*}
And
\begin{align}
y_c = c_1e^{-t}\cos(2t) + c_2e^{-t}\sin(2t)
\end{align}
Now solve non homogenous part
\begin{align*}
y'' + 2y' + 5y &= 8e^{-t}\\
\end{align*}
Let $Y_1 = Ae^{-t}$, then $Y_1' = -Ae^{-t},\ Y_1'' = Ae^{-t}$
\begin{align*}
A - 2A + 5A &= 8\\
A &= 2\\
Y_1 = 2e^{-t}
\end{align*}
And
\begin{align*}
y'' + 2y' + 5y &= 34\sin(2t)\\
\end{align*}
Let
\begin{align*}
Y_2 &= B\cos(2t) + C\sin(2t)\\
Y_2' &= 2C\cos(2t) - 2B\sin(2t)\\
Y_2'' &= -4B\cos(2t) - 4C\sin(2t)\\
\end{align*}
and
\begin{align*}
-4B\cos(2t) - 4C\sin(2t) + 4C\cos(2t) - 4B\sin(2t) + 5B\cos(2t) + 5C\sin(2t) &= 34\sin(2t)\\
B\cos(2t) + C\sin(2t) + 4C\cos(2t) - 4B\sin(2t) &= 34\sin(2t)
\end{align*}
Solve the linear equation
\begin{align*}
B+4C&=0\\
C-4B&=34\\
B = -8 &,\ C=2\\
Y_2 = 2\sin(2t)&-8\cos(2t)
\end{align*}
Combining all we have
\begin{align*}
y = c_1e^{-t}\cos(2t) + c_2e^{-t}\sin(2t) + 2e^{-t} + 2\sin(2t)-8\cos(2t)
\end{align*}
First solve homogenous complement equation
\begin{align*}
y'' + 2y' + 5y &= 0\\
r^2 + 2r + 5 &= 0\\
r_1 = 1+2i,\ &r_2 = 1-2i
\end{align*}
And
\begin{align}
y_c = c_1e^{-t}\cos(2t) + c_2e^{-t}\sin(2t)
\end{align}
Now solve non homogenous part
\begin{align*}
y'' + 2y' + 5y &= 8e^{-t}\\
\end{align*}
Let $Y_1 = Ae^{-t}$, then $Y_1' = -Ae^{-t},\ Y_1'' = Ae^{-t}$
\begin{align*}
A - 2A + 5A &= 8\\
A &= 2\\
Y_1 = 2e^{-t}
\end{align*}
And
\begin{align*}
y'' + 2y' + 5y &= 34\sin(2t)\\
\end{align*}
Let
\begin{align*}
Y_2 &= B\cos(2t) + C\sin(2t)\\
Y_2' &= 2C\cos(2t) - 2B\sin(2t)\\
Y_2'' &= -4B\cos(2t) - 4C\sin(2t)\\
\end{align*}
and
\begin{align*}
-4B\cos(2t) - 4C\sin(2t) + 4C\cos(2t) - 4B\sin(2t) + 5B\cos(2t) + 5C\sin(2t) &= 34\sin(2t)\\
B\cos(2t) + C\sin(2t) + 4C\cos(2t) - 4B\sin(2t) &= 34\sin(2t)
\end{align*}
Solve the linear equation
\begin{align*}
B+4C&=0\\
C-4B&=34\\
B = -8 &,\ C=2\\
Y_2 = 2\sin(2t)&-8\cos(2t)
\end{align*}
Combining all we have
\begin{align*}
y = c_1e^{-t}\cos(2t) + c_2e^{-t}\sin(2t) + 2e^{-t} + 2\sin(2t)-8\cos(2t)
\end{align*}
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Term Test 1 / Re: TT1 Problem 3 (main)
« on: October 16, 2018, 08:58:59 AM »
a). First solve the homogenous part
\begin{align*}
y'' + 5y' + 6y &= 0\\
r^2 + 5r + 6 &=0\\
(r+2)(r+3) &=0\\
r_1 = -2,\ r_2 &= -3
\end{align*}
So solution to homogenous part is
\begin{align*}
y_c = c_1e^{-2t} + c_2e^{-3t}
\end{align*}
Next we solve \begin{align}
y'' + 5y' + 6y &= -30e^{2t}
\end{align}
Let
\begin{align*}
Y_1 &= Ae^{2t}, \text{and}\\
Y_1' &= 2Ae^{2t}\\
Y_1'' &= 4Ae^{2t}, \text{Combining with (1)}
\end{align*}
\begin{align*}
4Ae^{2t} + 10Ae^{2t} &+ 6Ae^{2t} = -30e^{2t}\\
A &= -\frac{3}{2}\\
Y_1 &= -\frac{3}{2}e^{2t}
\end{align*}
Next we solve
\begin{align}
y'' + 5y' + 6y &= 3e^{-2t}
\end{align}
Since we already have $e^{-2t}$ in our solution, let
\begin{align*}
Y_2 &= Bte^{-2t}, \text{and}\\
Y_2' &= Be^{-2t} - 2Bte^{-2t}\\
Y_2'' &= -4Be^{-2t} + 4Bte^{-2t}, \text{Combining with (2)}
\end{align*}
\begin{align*}
(-4Be^{-2t} + 4Bte^{-2t}) + &5(Be^{-2t} - 2Bte^{-2t}) + 6(Bte^{-2t}) = 3e^{-2t}\\
B &= 3\\
Y_2 &=3te^{-2t}
\end{align*}
Combining above we have the general solution
\begin{align}
y &= c_1e^{-2t} + c_2e^{-3t} -\frac{3}{2}e^{2t} + 3te^{-2t}\\
y' &= -2c_1e^{-2t} - 3c_2e^{-3t} - 3e^{2t} +3e^{-2t} - 6te^{-2t}
\end{align}
b). Plug in $y(0)=0, y'(0) = 0$ to (3) and (4)
with
\begin{align*}
c_1 + c_2 &= \frac{3}{2}\\
-2c_1 - 3c_2 &= 0\\
c_1 = \frac{9}{2}&,\ c_2 = -3
\end{align*}
And finally,
\begin{align}
y &= \frac{9}{2}e^{-2t} - 3e^{-3t} -\frac{3}{2}e^{2t} + 3te^{-2t}\\
\end{align}
\begin{align*}
y'' + 5y' + 6y &= 0\\
r^2 + 5r + 6 &=0\\
(r+2)(r+3) &=0\\
r_1 = -2,\ r_2 &= -3
\end{align*}
So solution to homogenous part is
\begin{align*}
y_c = c_1e^{-2t} + c_2e^{-3t}
\end{align*}
Next we solve \begin{align}
y'' + 5y' + 6y &= -30e^{2t}
\end{align}
Let
\begin{align*}
Y_1 &= Ae^{2t}, \text{and}\\
Y_1' &= 2Ae^{2t}\\
Y_1'' &= 4Ae^{2t}, \text{Combining with (1)}
\end{align*}
\begin{align*}
4Ae^{2t} + 10Ae^{2t} &+ 6Ae^{2t} = -30e^{2t}\\
A &= -\frac{3}{2}\\
Y_1 &= -\frac{3}{2}e^{2t}
\end{align*}
Next we solve
\begin{align}
y'' + 5y' + 6y &= 3e^{-2t}
\end{align}
Since we already have $e^{-2t}$ in our solution, let
\begin{align*}
Y_2 &= Bte^{-2t}, \text{and}\\
Y_2' &= Be^{-2t} - 2Bte^{-2t}\\
Y_2'' &= -4Be^{-2t} + 4Bte^{-2t}, \text{Combining with (2)}
\end{align*}
\begin{align*}
(-4Be^{-2t} + 4Bte^{-2t}) + &5(Be^{-2t} - 2Bte^{-2t}) + 6(Bte^{-2t}) = 3e^{-2t}\\
B &= 3\\
Y_2 &=3te^{-2t}
\end{align*}
Combining above we have the general solution
\begin{align}
y &= c_1e^{-2t} + c_2e^{-3t} -\frac{3}{2}e^{2t} + 3te^{-2t}\\
y' &= -2c_1e^{-2t} - 3c_2e^{-3t} - 3e^{2t} +3e^{-2t} - 6te^{-2t}
\end{align}
b). Plug in $y(0)=0, y'(0) = 0$ to (3) and (4)
with
\begin{align*}
c_1 + c_2 &= \frac{3}{2}\\
-2c_1 - 3c_2 &= 0\\
c_1 = \frac{9}{2}&,\ c_2 = -3
\end{align*}
And finally,
\begin{align}
y &= \frac{9}{2}e^{-2t} - 3e^{-3t} -\frac{3}{2}e^{2t} + 3te^{-2t}\\
\end{align}
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MAT244--Lectures & Home Assignments / Typo in 2.5 Q23
« on: October 05, 2018, 01:11:41 AM »
Hello, professor, there is a typo that reads
$\frac{x}{dt} = âˆ′\alpha xy$
Should be
$\frac{dx}{dt} = -\alpha xy$
Also could you fix the link of "Test 1" on course page? It links to last year's.
$\frac{x}{dt} = âˆ′\alpha xy$
Should be
$\frac{dx}{dt} = -\alpha xy$
Also could you fix the link of "Test 1" on course page? It links to last year's.
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MAT244--Lectures & Home Assignments / Re: question about the acid water problem
« on: September 27, 2018, 10:10:24 AM »
You could read the code to see the solution, or I'm sure he'll tell you if you email to ask.
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