Messages

1

Final Exam / Re: FE-P1

« on: December 14, 2018, 12:25:45 PM »

$M = 2x \sin(y) +1 $

$N = 4 x^2 \cos(y) + 3x \cot(y) + 5 \sin(2y) $

$M_y = 2x\cos(y)$

$N_x = 8x\cos(y) + 3cos(y)$

$\frac{M_y - N_x} {M} = \frac{-6x\cos(y) - 3cos(y)}{2x\sin(y) + 1} = -3\cot(y)$

$\mu = e^{-\int -3\cot(y) dy } = \sin^3(y)$

Then multiply $sin^3(y)$ on M and N.

Then we get $M =  sin^3(y)(2x \sin(y) +1)$, $N =sin^3(y)(4 x^2 \cos(y) + 3x \cot(y) + 5 \sin(2y))$

$M_y = \sin^2(y) \cos(y) (8x \sin(y) + 3) $, and $N_x = \sin^2(y) \cos(y) (8x \sin(y) + 3) $

Thus $M_y = N_x$, exact.

Then $\psi_x = M$

$\psi = \int M dx = \int \sin^3(y)(2x \sin(y) +1) dx = \sin^3(y)(x^2\sin(y) + x) + h(y) $

$N = \psi_y = 4x^2\sin^3(y)\cos(y) + x\sin^2(y)\cos(y) + h'(y) $

$h'(y) = 10\sin^4cos(y) $

$h(y) = \int 10\sin^4cos(y) dy = 2\sin^5(y) + c$

Then $\psi = \sin^3(y)(x^2\sin(y) + x) + 2\sin^5(y) = c$

2

MAT244--Misc / Quiz 6 mark

« on: December 05, 2018, 03:51:23 PM »

When can we get the mark of quiz 6?

3

MAT244--Lectures & Home Assignments / Re: Final review of 9.1

« on: November 27, 2018, 06:09:55 PM »

\begin{equation*}
   \begin{bmatrix}
    -1 & -1 \\
    2 & -1
    \end{bmatrix}x^0 +
    \begin{bmatrix}
    -1 \\ 5
    \end{bmatrix} = 0
\end{equation*}
\begin{equation*}
     \begin{bmatrix}
    -1 & -1 \\
    2 & -1
    \end{bmatrix}x^0 = -\begin{bmatrix}
    -1 \\ 5
    \end{bmatrix}
\end{equation*}
\begin{align*}
    x_0 &=  \begin{bmatrix}
    -1 & -1 \\
    2 & -1
    \end{bmatrix}^{-1}\begin{bmatrix}
    1 \\ -5
    \end{bmatrix}\\ &=
    \begin{bmatrix}
    -2 \\ 1
    \end{bmatrix}
\end{align*}
Hence the critical point is $x_0 = -2$ and $y_0 = 1$

Let $x = x^0 + u$

Then $\frac{dx}{dt} = \frac{du}{dt}$ (as $\frac{dx^0}{dt} = 0$)

Then \begin{align*}
    \frac{du}{dt} &= \begin{bmatrix}
    -1 & -1 \\
    2 & -1
    \end{bmatrix}\begin{bmatrix}
    -2 \\ 1
    \end{bmatrix}  + \begin{bmatrix}
    -1 & -1 \\
    2 & -1
    \end{bmatrix}u + \begin{bmatrix}
    -1 \\ 5
    \end{bmatrix} \\
    &= -\begin{bmatrix}
    -1 \\ 5
    \end{bmatrix} +
    \begin{bmatrix}
    -1 & -1 \\
    2 & -1
    \end{bmatrix}u + \begin{bmatrix}
    -1 \\ 5
    \end{bmatrix}
\end{align*}
\begin{equation*}
    \frac{du}{dt} = \begin{bmatrix}
    -1 & -1 \\
    2 & -1
    \end{bmatrix}u
\end{equation*}

Let $u = ae^{rt} $
\begin{equation*}
   \begin{bmatrix}
    -1-r & -1 \\
    2 & -1-r
    \end{bmatrix}
    \begin{bmatrix}
    a_1 \\ a_2
    \end{bmatrix} =
    \begin{bmatrix}
    0 \\ 0
    \end{bmatrix}
\end{equation*}
\begin{equation*}
    r^2 + 2r + 3 = 0
\end{equation*}
\begin{equation*}
    r = -1 + \pm \sqrt{2}i
\end{equation*}

Thus the eigenvalue are $-1 + \pm \sqrt{2}i$, the critical point is asymptotically stable spiral point

4

MAT244--Lectures & Home Assignments / Re: Final review question

« on: November 27, 2018, 05:18:44 PM »

$r^3 - 2r^2 + 4r - 8 = 0$

$(r-2)r^2 + 4(r-2) = 0$

$(r-2)(r^2+4) = 0$

r =2, r =$\pm$2i

Then

$y_c(t) = c_1e^{2t} + c_2\cos2t + c_3\sin2t$

$y^{'''} - 2y^{''} + 4y' - 8y = 16e^{2t}$

$y_p(t) = Ae^{2t} = Ate^{2t}$

$(Ate^{2t})^{'''} - 2(Ate^{2t})^{''} + 4(Ate^{2t})' - 8(Ate^{2t}) = 16e^{2t}$

Then we get A = 2

Then $y_p(t) = 2e^{2t}$

$y^{'''} - 2y^{''} + 4y' - 8y = 30\cos t$

$y_p(t) = B\cos t + C\sin t$

$(B\cos t + C\sin t)^{'''} - 2(B\cos t + C\sin t)^{''} + 4(B\cos t + C\sin t)' - 8(B\cos t + C\sin t) = 30\cos t$

Then we get B=-4 and C=2

$y_p(t) = 2 \sin t - 4 \cos t$

Thus, $y(t) = c_1e^{2t} +c_2\cos(2t)+c_3\sin(2t)+2te^t+2\sin t - 4\cos t$

5

MAT244--Lectures & Home Assignments / Re: 9.2 problem set Q21?

« on: November 24, 2018, 05:13:09 PM »

Since $x=\phi(t)$, $y=\psi(t)$ is a solution of the autonomous system
\begin{equation}
    \frac{dx}{dt} = F(x,y), \frac{dy}{dt}=G(x,y)
\end{equation}
for $\alpha < t < \beta$.

Then functions  $\Phi(t)=\phi(t-s)$ and $\Psi(t)=\psi(t-s)$ indicates same function $x=\phi(t)$ and $y=\psi(t)$ respectively (except time shift by a real number s)
Base on the definition of Autonomous system (on textbook page 509) "an autonomous system is one whose configuration, including physical parameters and external forces or effects, is independent of time."

Thus, $x = \Phi(t)=\phi(t-s)$, $y=\Psi(t)=\psi(t-s)$ is a solution for $\alpha +s < t < \beta$ +s for any real number s.

6

Quiz-5 / Re: Q5 TUT 5102

« on: November 03, 2018, 01:33:22 AM »

thank you, fix it

7

Quiz-5 / Re: Q5 TUT 5101

« on: November 02, 2018, 04:15:18 PM »

$$\tan(z) = \frac{\sin(z)}{\cos (z)} = \frac{z - \frac{1}{3!}z^3 + \frac{z^5}{5!} - \frac{z^7}{7!}...}{1 - \frac{z^2}{2!}+\frac{z^4}{4!}...} = a_0 + a_1z + a_2z^2 +a_3z^3 + ...$$

Thus
$(a_0 + a_1z + a_2z^2 +a_3z^3 + ...)(1 - \frac{z^2}{2!}+\frac{z^4}{4!}...) = z - \frac{1}{3!}z^3 + \frac{z^5}{5!} - \frac{z^7}{7!}...$

Thus we get

$a_0 = 0$

$a_1 = 1$

$a_2 = 0$

.
.
.

Therefore $\tan(z) = z + \frac{1}{3}z^3 + \frac{2}{15}z^5 ....$

8

Quiz-5 / Re: Q5 TUT 5301

« on: November 02, 2018, 03:43:00 PM »

\begin{align*}
 \frac{z+2}{z+3} &= \frac{z+3-1}{z+3} \\ &= 1-\frac{1}{z+3}\\
 &= 1- \frac{1}{z+1+2} \\ &= 1 - \frac{1}{2} \frac{1}{1+\frac{z+1}{2}}\\
 &= 1 - \frac{1}{2} \frac{1}{1-\frac{-(z+1)}{2}}\\
 &= 1 - \sum_{n=0}^{\infty}(\frac{-(z+1)}{2})^n \\
 &= 1 - \sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}(z-(-1))^n
\end{align*}

9

Quiz-5 / Re: Q5 TUT 5201

« on: November 02, 2018, 03:41:27 PM »

\begin{align*}
 \frac{z+2}{z+3} &= \frac{z+3-1}{z+3} \\ &= 1-\frac{1}{z+3}\\
 &= 1- \frac{1}{z+1+2}\\ &= 1 - \frac{1}{2} \frac{1}{1+\frac{z+1}{2}}\\
 &= 1 - \frac{1}{2} \frac{1}{1-\frac{-(z+1)}{2}}\\
 &= 1 - \sum_{n=0}^{\infty}(\frac{-(z+1)}{2})^n \\
 &= 1 - \sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}(z-(-1))^n
\end{align*}

10

Quiz-5 / Re: Q5 TUT 0101

« on: November 02, 2018, 03:29:12 PM »

Let $z^2(1-\cos(z)) = 0$

Then $z = 0$ or $\cos(z) = 1$

thus $z = 2k\pi$

 

case $1$: $z = 0$

let $f(z) = z^2$ and $h(z) = 1-\cos(z)$

Then $f(0)=0$

$f^{'}(0) = 2z|_{z=0} = 0$

$f^{''}(0)=2\neq 0$

thus order = 2

Then $h(0) = 0$

$h^{'}(0) = 2\sin(z)|_{z=0} = 0$

$f^{''}(0)=\cos(z)\neq 0$

thus order = 2

Therefore order(0) = 4



case $2$: $z = 2k\pi$ ($k\neq 0)$

let $f(z) = z^2$ and $h(z) = 1-\cos(z)$

Then $f(z) = z^2 = (2k\pi)^2 \neq 0$

thus order = 0

$h(z) = 1- \cos(z) = 0 $

$h^{'}(z) = \sin(z) = 0$

$h^{''} = \cos(z) = 0$

thus order = 2

Therefore order($2k\pi$) = 2, $k\neq0$

11

Quiz-5 / Re: Q5 TUT 5102

« on: November 02, 2018, 03:20:04 PM »

homogeneous equation is
$y^{'''} - y^{'} = 0$

then $r^3 - r = 0$

$r = 0, \pm 1$

thus, $y_{c}(t) = c_{1} + c_{2}e^t + c_{3}e^{-t}$

\begin{align*}
W(t) &=

\begin{bmatrix}
1 & e^t & e^{-t} \\
0 & e^t & -e^{-t}\\
0 & e^t & e^{-t}
\end{bmatrix} \\
\\

&= 1*(-1)^{1+1}
\begin{bmatrix}
e^t & -e^{-t}\\
e^t & e^{-t}
\end{bmatrix}
+ e^t(-1)^{2+1}
\begin{bmatrix}
0 & -e^{-t}\\
0 & e^{-t}
\end{bmatrix}
+ e^{-t}(-1)^{3+1}
\begin{bmatrix}
0 & e^t\\
0 & e^t
\end{bmatrix}\\
\\

&= e^t*e^t - (-e^{-t})e^t + 0 + 0 \\
\\

&= 1 + 1 \\
\\

&= 2
\end{align*}

\begin{align*}
W_{1}(t) &=
\begin{bmatrix}
0 & e^t & e^{-t}\\
0 & e^t & -e^{-t}\\
1 & e^t & e^{-t}
\end{bmatrix}\\
\\

&=1*(-1)^{1+3}
\begin{bmatrix}
e^t & e^{-t}\\
e^t & -e^{-t}
\end{bmatrix}\\
\\

&= -1-1 \\
\\

&= -2
\end{align*}

\begin{align*}
W_{2}(t) &=
\begin{bmatrix}
1 & 0 & e^{-t}\\
0 & 0 & -e^{-t}\\
0 & 1 & e^{-t}
\end{bmatrix}\\
\\

&=1*(-1)^{1+1}
\begin{bmatrix}
0 & -e^{-t}\\
1 & e^{-t}
\end{bmatrix}\\
\\

&= e^{-t}
\end{align*}
\\

\begin{align*}
W_{3}(t) &=
\begin{bmatrix}
1 & e^t & 0\\
0 & e^t & 0\\
0 & e^t & 1
\end{bmatrix}\\
\\

&=1*(-1)^{1+1}
\begin{bmatrix}
e^{t} & 0\\
e^{t} & 1
\end{bmatrix}\\
\\

&= e^{t}
\end{align*}

\begin{align*}
y_{p}(t) &= y_{1}\int \frac{g(t)W_{1}(t)}{W(t)} dt + y_{2}\int \frac{g(t)W_{2}(t)}{W(t)} dt + y_{3}\int \frac{g(t)W_{3}(t)}{W(t)} dt \\
\\

&= 1\int \frac{\cosh(t)(-2)}{2)} dt + e^t\int \frac{\cosh(t)e^{-t}}{2} dt + e^{-t}\int \frac{\cosh(t)e^t}{2} dt \\
\\

&= -1\sinh(t) + \frac{1}{2}e^t\int \cosh(t)e^{-t} dt + \frac{1}{2}e^{-t}\int \cosh(t) e^t dt \\
\\

&= -\sinh(t) + \frac{1}{2}e^t \int \cosh(t)e^{-t}dt + \frac{1}{2}e^{-t}\int \cosh(t)e^t dt \\
\\

&= -\sinh(t) + \frac{1}{2}e^t(\frac{t}{2} - \frac{e^{-2t}}{4}) + \frac{1}{2}e^{-t}(\frac{t}{2}+ \frac{e^{2t}}{4})
\end{align*}

Therefore $y(t) = c_{1} + c_{2}e^t + c_{3}e^{-t} -\sinh(t) + \frac{1}{2}e^t(\frac{t}{2} - \frac{e^{-2t}}{4}) + \frac{1}{2}e^{-t}(\frac{t}{2}+ \frac{e^{2t}}{4})$

12

Quiz-5 / Re: Q5 TUT 0401

« on: November 02, 2018, 03:19:43 PM »

The homogeneous equation is $y^{'''} + y^{'} = 0$

$r^3 + r = 0$

$r = 0, \pm i$

Thus the roots are $r = 0, \pm i$

Thus $y_{c} = c_1 + c_2\cos(t) + c_3\sin(t)$

\begin{equation*}
W(t) =
W(1, \cos (t), \sin (t))
= \begin{bmatrix}
1 & \cos(t) & \sin(t) \\
0 & -\sin(t) & \cos(t) \\
0 & -\cos(t) &-\sin(t)
\end{bmatrix}
= \sin^{2}(t) + \cos^{2} (t)
= 1
\end{equation*}

\begin{equation*}
W_{1}(t) =
\begin{bmatrix}
0 & \cos(t) & \sin(t) \\
0 & -\sin(t) & \cos(t) \\
1 & -\cos(t) &-\sin(t)
\end{bmatrix}
= \sin^{2}(t) + \cos^{2} (t)
= 1
\end{equation*}

\begin{equation*}
W_{2}(t) =
\begin{bmatrix}
1 & 0 & \sin(t) \\
0 & 0 & \cos(t) \\
1 & 1 &-\sin(t)
\end{bmatrix}
= -\cos (t)
\end{equation*}

\begin{equation*}
W_{3}(t) =
\begin{bmatrix}
1 & \cos (t) & 0 \\
0 & -\sin(t) & 0 \\
0 & -\cos(t) & 1
\end{bmatrix}
= -\sin (t)
\end{equation*}

\begin{align*}
y_{p}(t) &= y_{1}\int \frac{g(t)W_{1}(t)}{W(t)} dt + y_{2}\int \frac{g(t)W_{2}(t)}{W(t)} dt + y_{3}\int \frac{g(t)W_{3}(t)}{W(t)} dt \\
\\
&= 1\int \frac{\sec(t) 1}{1} dt + \cos (t)\int \frac{\sec (t) (-\cos(t))}{1} dt + \sin(t)\int \frac{\sec(t)(-\sin(t))}{1} dt\\
\\
&= \ln (\sec(t)+\tan(t))-t\cos(t) + \sin(t)\ln(\cos(t))
\end{align*}

Thus, the general solution is

$y(t) = c_1 + c_2\cos(t) + c_3\sin(t) + \ln(\sec(t)+\tan(t))-t\cos(t)+\sin(t)\ln(\cos(t))$

13

MAT244--Lectures & Home Assignments / Re: section 7.3 question 24

« on: November 01, 2018, 01:18:46 PM »

\begin{equation*}
A=
\begin{bmatrix}
11 & -2 & 8\\ -2 & 2 & 10 \\ 8 & 10 & 5
\end{bmatrix}
\end{equation*}
then
\begin{equation*}
(A -\lambda I)x = 0
\end{equation*}
then
\begin{equation*}
\begin{bmatrix}
11-\lambda & -2 & 8\\ -2 & 2-\lambda & 10 \\ 8 & 10 & 5-\lambda
\end{bmatrix}
\begin{bmatrix}
x_1 \\ x_2 \\x_3
\end{bmatrix} =
\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}
\end{equation*}
then the eigenvalues are the roots of equation det(A-$\lambda$ I) = 0
\begin{equation*}
\left|\begin{matrix}
11-\lambda & -2 & 8\\ -2 & 2-\lambda & 10 \\ 8 & 10 & 5-\lambda
\end{matrix}\right| = 0 \qquad \tag{*}
\end{equation*}
$C_2 -> 4C_2 + C_3$
\begin{equation*}
\begin{bmatrix}
11-\lambda & 4(-2)+8 & 8\\ -2 & 4(2-\lambda)+10 & 10 \\ 8 & 4(10)+(5-\lambda) & 5-\lambda
\end{bmatrix} = 0
\end{equation*}
\begin{equation*}
\begin{bmatrix}
11-\lambda &0 & 8\\ -2 & 18-4\lambda & 10 \\ 8 & 45-\lambda & 5-\lambda
\end{bmatrix} = 0
\end{equation*}
\begin{equation*}
(11-\lambda)
\begin{bmatrix}
18-4\lambda & 10 \\ 45-\lambda & 5-\lambda
\end{bmatrix} - 0
\begin{bmatrix}
-2 & 10 \\ 8 & 5-\lambda
\end{bmatrix} + 8
\begin{bmatrix}
-2 & 18-4\lambda \\ 8 & 45 -\lambda
\end{bmatrix} = 0
\end{equation*}

(11-$\lambda$)(18-4$\lambda$)(5-$\lambda$)(10)(45-$\lambda$)+8(-2)(45-$\lambda$)8(18-4$\lambda$)=0
then -4$\lambda^3$ + 8 $\lambda^2$ + 4$\lambda$ - 8 = 0
($\lambda$ -1 )($\lambda$ - 2) ($\lambda$ +1) = 0
then $\lambda$ = 1,2,-1
then the eigenvalues are $\lambda_1$ = 1, $\lambda_2$ = 2, $\lambda_3$ = -1

14

Thanksgiving bonus / Re: Thanksgiving bonus 4

« on: October 06, 2018, 10:58:53 PM »

sorry, there is a typo. The solution should be the given function doesn't represent a locally sourceless and an irrotational flow 

15

Thanksgiving bonus / Re: Thanksgiving bonus 4

« on: October 06, 2018, 10:32:20 PM »

$cosh(x)cos(y)+isinh(x)sin(y)$
Then
$u=cosh(x)cos(y)$
$v=sinh(x)sin(y)$
$\frac{du}{dx} = sinh(x)cos(y)$
$\frac{du}{dy}= -cosh(x)sin(y)$
$\frac{dv}{dx} = cosh(x)sin(y)$
$\frac{dv}{dy} = sinh(x)cos(y)$
since $\frac{du}{dx} \neq - \frac{dv}{dy}$
$\frac{du}{dy} \neq \frac{dv}{dx}$
Thus, the flow is not globally sourceless and is not irrotational.