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### Messages - Rhoda Lam

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1
##### MAT244 Misc / Re: Past Test 2
« on: November 18, 2014, 03:10:04 PM »

2
##### Quiz 3 / Q3 problem 2 (day section)
« on: October 28, 2014, 12:43:57 PM »
7.5 p 405,# 12
Solve the following equation.
\begin{equation*}
x' = \begin{pmatrix} 3 & 2 & 4 \\ 2 & 0 & 2 \\ 4 & 2 & 3 \end{pmatrix}x
\end{equation*}

First, find the eigenvalues.

\begin{equation*}
\begin{pmatrix} 3-\lambda & 2 & 4 \\2 &  -\lambda & 2 \\ 4 & 2 & 3-\lambda\end{pmatrix} = 0  \notag
\end{equation*}

The characteristic equation is:
\begin{gather}
\lambda^3 - 6\lambda^2 - 15\lambda - 8 = 0\\
(1 + \lambda)( (1 + \lambda)(8 - \lambda) = 0\\
\end{gather}

Therefore, the eigenvalues are $\lambda = 8,\ \lambda = -1,\ \lambda = -1$.

For $\lambda = 8$:
\begin{equation*}
\begin{pmatrix}
-5 & 2 & 4
\\2 & -8 & 2
\\ 4 & 2 & -5\end{pmatrix}
\begin{pmatrix} x \\ y \\ z \end{pmatrix} =
\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\end{equation*}
The corresponding eigenvector is $v_1 = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}$

For $\lambda = -1$:
\begin{equation*}
\begin{pmatrix}
4 & 2 & 4
\\2 & 1 & 2
\\ 4 & 2 & 4\end{pmatrix}
\begin{pmatrix} x \\ y \\ z \end{pmatrix} =
\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\end{equation*}
The corresponding eigenvectors are $v_2 = \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}$ and $v_3 = \begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix}$

Therefore, the general solution is:

\begin{equation*}x = c_1\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}e^{8t} + c_2\begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}e^{-t} + c_3\begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix}e^{-t} \end{equation*}

3
##### Quiz 1 / Re: Q1 problem 2 (day section)
« on: October 05, 2014, 09:23:47 PM »
2.6 p. 101, #22
Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.

(x + 2) \sin y + (x \cos y)y' = 0\label{A}\\
Î¼(x, y) = xe^x

Let $M(x,y) = (x+2)\sin y$, $N(x,y) = x(\cos y)$. Then $M_y(x,y) = (x+2)\cos y$, $N_x(x,y) = \cos y$.

Since $M_y(x,y) \ne N_x(x,y)$, then equation (\ref{A}) is not exact. Multiply the given integrating factor to Equation (\ref{A}):

xe^x(x + 2) \sin y + xe^x(x \cos y)y' = 0\label{B}\\

Now $M(x,y) = xe^x(x+2)\sin y$, $N(x,y) = x^2e^x(\cos y)$

Then $M_y(x,y) = xe^x(x+2)\cos y$, $N_x(x,y) = 2xe^x\cos y + x^2e^x cos y = xe^x(x+2) \cos y$. Since $M_y(x,y) = N_x(x,y)$, then Equation (\ref{B}) is exact.

There is a  $\Psi(x, y)$ such that:
\begin{gather}
\Psi _x(x, y) = M(x,y) = xe^x(x+2)\sin y\label{C}\\
\Psi _y (x, y) = N(x,y) = x^2e^x(\cos y)\label{D}
\end{gather}

Integrate Equation (\ref{C}) and we get $\Psi (x, y) =\sin(y)x^2e^x + g(y)$.

Now differentiate it to get $\Psi_y (x, y) = \cos(y)x^2e^x + g'(y) = \cos(y)x^2e^x \implies g'(y) = 0 \implies g(y) = C$

Therefore, the solution is:

x^2e^x\sin(y) = C

Observe I write \cos, \sin , \ln etc, to keep them upright rather than italic and have a proper spacing; in $\LaTeX$ they are called operators-V.I.

4
##### Quiz 1 / Re: Q1 problem 1 (day section)
« on: October 04, 2014, 09:42:22 PM »
2.1 p. 40, #18
\begin{gather}
ty' + 2y = \sin t,\\
y(Ï€/2) = 1
\end{gather}
First, change the equation by dividing every term by $t$.

y' + (2/t)y = (\sin (t))/t

Then, solve for the integrating factor.

Î¼ = e^{\int (2/t)\,dt} = e^{2\ln(t)\,} = t^2

Multiply every term in equation by Î¼.

t^2y' + 2ty = t\sin(t)

Integrate both sides of the equation.
\begin{gather}
\int[t^2y]' = \int{t\sin(t)}dt\implies
t^2y = âˆ’t\cos(t) âˆ’ \int{-\cos(t)}dt= âˆ’t\cos(t) + \sin(t) + C\implies y = (âˆ’t \cos(t) +\sin (t) + C)/t^2
\end{gather}
Substitute the initial value to solve for C.
\begin{gather}
C = (Ï€/2)^2-1
\end{gather}
Therefore, the solution is:
\begin{gather}
y = (âˆ’t \cos(t) +\sin (t) + (Ï€/2)^2-1)/t^2
\end{gather}

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