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**MAT334--Lectures & Home Assignments / Re: 2.5 - Q23**

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**Today**at 05:11:49 AM »

**Ende**

Fix parenthesis

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Fix parenthesis

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Search and read before you write

http://forum.math.toronto.edu/index.php?topic=1242.0

http://forum.math.toronto.edu/index.php?topic=1242.0

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TT2 went rather smoothly, but with some exceptions

1) Very few students disobeyed instructions "Stop writing". On FE CPO (Chief Presiding Officer, aka Dean's representative) will be doing with such issues, and there will be rather serious repercussions.

2) Not all students followed sitting instructions and it made signature collection longer and more complicated. Please follow these instructions on FE, which will be exactly in the same rooms. However, without early sittings, approximately 40 extra students will be in each room. CPO will collect signatures.

3) Someone wrote a part of the solution on the provided scrap paper. When this was discovered, I allowed to copy one-to-one to the booklet immediately after exam, since Crowdmark grading of anything which is not in the booklet is almost impossible; so without some efforts this solution would be lost. I am not sure if CPO would allow such resque. Remember: scrap paper is not submitted, and if submitted, just discarded.

1) Very few students disobeyed instructions "Stop writing". On FE CPO (Chief Presiding Officer, aka Dean's representative) will be doing with such issues, and there will be rather serious repercussions.

2) Not all students followed sitting instructions and it made signature collection longer and more complicated. Please follow these instructions on FE, which will be exactly in the same rooms. However, without early sittings, approximately 40 extra students will be in each room. CPO will collect signatures.

3) Someone wrote a part of the solution on the provided scrap paper. When this was discovered, I allowed to copy one-to-one to the booklet immediately after exam, since Crowdmark grading of anything which is not in the booklet is almost impossible; so without some efforts this solution would be lost. I am not sure if CPO would allow such resque. Remember: scrap paper is not submitted, and if submitted, just discarded.

5

When considering system

\begin{equation*}

\mathbf{x}'=A\mathbf{x} \qquad \text{with } \

A=\begin{pmatrix} a & b \\ c & d\end{pmatrix}

\end{equation*}

and discovering that it has a repeated root $\mu\ne 0$, but only one eigenvector, we know that $\mu>0$ corresponds to an unstable node, and $\mu<0$ t corresponds to a stable node. Furthermore, on the pictures below a straight line is directed along this single eigenvector. However, how to distinguish between two upper pictures and two lower pictures?

Observe that, $\mu$ is a double root of the equation $\lambda ^2- (a+d)\lambda + ad -bc=0$ with the discriminant $D:=(a+d)^2-4(ad -bc))=(a-d)^2+4bc$, and we consider the case $D=0\implies bc <0$ (except $a=d$). Thus $b$ and $c$ are not $0$ and have opposite signs.

Then, if $b<0$ (and $c>0$) "rotation" is counter-clockwise, and if $b>0$ (and $c<0$) "rotation" is clockwise. I say "rotation" because it is not a real rotation as in the case of complex-conjugate roots, only a half-turn rotation by $\pm \pi$, but it works! Indeed, if we slightly increase $|b|$ or $|c|$, we get $D<0$ and we will have a focal point and the picture needs to be consistent.

If $a=d=\mu$ then either $b=0$ or $c=0$ but not both (otherwise there would be 2 linearly independent eigenvectors) and we use the same sign criteria, looking at $b$ or $c$ which is not $0$.

\begin{equation*}

\mathbf{x}'=A\mathbf{x} \qquad \text{with } \

A=\begin{pmatrix} a & b \\ c & d\end{pmatrix}

\end{equation*}

and discovering that it has a repeated root $\mu\ne 0$, but only one eigenvector, we know that $\mu>0$ corresponds to an unstable node, and $\mu<0$ t corresponds to a stable node. Furthermore, on the pictures below a straight line is directed along this single eigenvector. However, how to distinguish between two upper pictures and two lower pictures?

Observe that, $\mu$ is a double root of the equation $\lambda ^2- (a+d)\lambda + ad -bc=0$ with the discriminant $D:=(a+d)^2-4(ad -bc))=(a-d)^2+4bc$, and we consider the case $D=0\implies bc <0$ (except $a=d$). Thus $b$ and $c$ are not $0$ and have opposite signs.

Then, if $b<0$ (and $c>0$) "rotation" is counter-clockwise, and if $b>0$ (and $c<0$) "rotation" is clockwise. I say "rotation" because it is not a real rotation as in the case of complex-conjugate roots, only a half-turn rotation by $\pm \pi$, but it works! Indeed, if we slightly increase $|b|$ or $|c|$, we get $D<0$ and we will have a focal point and the picture needs to be consistent.

If $a=d=\mu$ then either $b=0$ or $c=0$ but not both (otherwise there would be 2 linearly independent eigenvectors) and we use the same sign criteria, looking at $b$ or $c$ which is not $0$.

6

$\frac{1}{4^{n+1}}$ cannot be outside of the sum

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Do not post solutions to more than one parallel problems for different sittings, f.e. TT2-P1 and TT2B-P1.

On the other hand, you may post solution to TT2-P1 and participate in the discussion for TT2A-P1 and TT2B-P1.

On the other hand, you may post solution to TT2-P1 and participate in the discussion for TT2A-P1 and TT2B-P1.

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$$

\mathbf{x}'=\begin{pmatrix}

-3 &-2\\

\hphantom{-}5 &-5\end{pmatrix}\mathbf{x}.$$

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$$

\mathbf{x}'=\begin{pmatrix} -1 &\hphantom{-}1\\

-1 &-3\end{pmatrix}\mathbf{x}.$$

$$

\mathbf{x}'=\begin{pmatrix} -1 & \hphantom{-}1\\

-1 &-3\end{pmatrix}\mathbf{x} +

\begin{pmatrix} \frac{e^{-2t}}{t^2+1}\\

0\end{pmatrix},\qquad

\mathbf{x}(0)=\begin{pmatrix} 0 \\

0\end{pmatrix}.

$$

10

Consider equation

\begin{equation}

y'''-y'' +4y'-4y= 8\cos(2t).

\label{2-1}

\end{equation}

**(a)** Write a differential equation for Wronskian of $y_1,y_2,y_3$, which are solutions for homogeneous equation and solve it.

**(b)** Find fundamental system $\{y_1,y_2,y_3\}$ of solutions for homogeneous equation, and find their Wronskian. Compare with (a).

**(c)** Find the general solution of (\ref{2-1}).

\begin{equation}

y'''-y'' +4y'-4y= 8\cos(2t).

\label{2-1}

\end{equation}

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(a) Find the general solution of

\begin{equation*}

y''-5y'+6y= \frac{6e^{4t}}{e^{2t}+1}.

\end{equation*}

**(b)** Find solution, such that $y(0)=0$, $y'(0)=0$.

\begin{equation*}

y''-5y'+6y= \frac{6e^{4t}}{e^{2t}+1}.

\end{equation*}

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$$

\mathbf{x}'=\begin{pmatrix}

\hphantom{-}1 & \hphantom{-}2\\

-5 &-1\end{pmatrix}\mathbf{x}.$$

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$$

\mathbf{x}'=\begin{pmatrix} \ 4 & \ 1\\

-3 &0\end{pmatrix}\mathbf{x}.$$

$$

\mathbf{x}'=\begin{pmatrix}\hphantom{-}4 & \ 1\\

-3 &0\end{pmatrix}\mathbf{x} +

\begin{pmatrix} \hphantom{-}\frac{4e^{4t}}{e^t+1} \\

-\frac{4e^{4t}}{e^t+1}\end{pmatrix},\qquad

\mathbf{x}(0)=\begin{pmatrix}-1 \\

\hphantom{-}3\end{pmatrix}.

$$

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Consider equation

\begin{equation}

y'''-2y'' -y'+2y= 8e^{t}.

\label{2-1}

\end{equation}

**(a)**

Write a differential equation for the Wronskian of $y_1,y_2,y_3$, which are solutions for homogeneous equation and solve it.

**(b)** Find fundamental system $\{y_1,y_2,y_3\}$ of solutions for homogeneous equation, and find their Wronskian. Compare with (a).

**(c)** Find the general solution of (\ref{2-1}).

\begin{equation}

y'''-2y'' -y'+2y= 8e^{t}.

\label{2-1}

\end{equation}

Write a differential equation for the Wronskian of $y_1,y_2,y_3$, which are solutions for homogeneous equation and solve it.

15

\begin{equation*}

y''-4y=2\tanh (t).

\end{equation*}