MATH7021: Spring 2018, Week 12

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Week 12

On Monday we finished the module by looking at triple integrals.

The Wednesday 09:00 lecture will be a tutorial. In this class and your usual tutorial we will look at the P.182, P. 192, P. 163, & P.116 exercises. If these are completed you will be recommended to revise either by trying Chapter 1 & 2 exercises or perhaps by looking at the Summer 2017 paper.

As next Monday is a bank holiday, we will begin the Summer 2017 Paper (in your notes) revision on Thursday.

Week 13

In the Wednesday 09:00 lecture we will continue working on the Summer 2017 Paper and hopefully finish it before the end of the Thursday 10:00 lecture.

If we finish the Summer 2017 paper early, any extra time (probably just Thursday but maybe Wednesday if we go fast) will be dedicated to one-to-one help.

Wednesday’s tutorial will go ahead as normal with one-to-one help.

Study

Please feel free to ask me questions about the exercises via email or even better on this webpage.

Student Resources

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May 9, 2018 at 12:30 pm

Student

I’m doing question 2, 2017 in the past exam and I don’t know if the answer is correct, could please check it. I send a copy of my solution.

May 9, 2018 at 1:59 pm

J.P. McCarthy

I’m afraid your solution is incorrect in many ways.

Looking at the auxiliary equation

$r^2+8r+7=0$

is correct. However the factors are

$(r+7)(r+1)=0 \Rightarrow r=-7,-1$ .

This gives a homogeneous solution:

$y_H(t)=A e^{-7t}+Be^{-1t}$ .

Now we look for a particular solution. We trial $y_T(t)=Ce^{-2t}$ . This has first and second derivatives $y'_T(t)=-2Ce^{-2t}$ and $y''_T(t)=+4Ce^{-2t}$ . Force this into the differential equation

$y''_T(t)+8y'_T(t)+7y_T(t)=4Ce^{-2t}+8(-2Ce^{-2t})+7Ce^{-2t}=-5Ce^{-2t}$ .

To satisfy the differential equation this must equal $10e^{-2t}$ :

$-5Ce^{-2t}\overset{!}{=}10e^{-2t}\Rightarrow C=-2$ , and so we have a particular solution

$y_P(t)=-2e^{-2t}$ .

So that the general solution is

$y_G(t)=A e^{-7t}+Be^{-1t}-2e^{-2t}$ .

Now we need to find $A$ and $B$ such that the initial conditions are satisfied. I will leave this to you.

Regards,
J.P.

May 10, 2018 at 12:05 pm

How do you spot that $s^2-0.01=(s-0.1)(s+0.1)$ .

Is there a long way of doing this out? I won’t get this in the exam if something like this comes up. Do you just have to know that $0.01$ changes to $s-0.1$ and $s+0.1$ ?

May 10, 2018 at 12:23 pm

There are quite a number of ways of seeing this.

It is probably worth pointing out a different problem. Suppose you have

$\displaystyle X(s)=\frac{s}{s^2+7}$ ,

and you want to send this back. You look in the table and see that

$\displaystyle \mathcal{L}^{-1}\left\{\frac{s}{s^2+\omega^2}\right\}$ but $7$ isn’t a square… or is it? Any positive number can be written as a square via $\displaystyle a=(\sqrt{a})^2$ , and so,

$\displaystyle X(s)=\frac{s}{s^2+(\sqrt{7})^2}\overset{\mathcal{L}^{-1}}{\longrightarrow} \cos(\sqrt{7}\cdot t)$ .

This is how you could spot that $0.01=0.1^2$ as $\sqrt{0.01}=0.1$ .

OK, I have four ways of getting $s^2-0.01=(s-0.1)(s+0.1)$ .

1. First of all, just by noticing that it is a difference of squares. We have

$a^2-b^2=(a-b)(a+b)$ ,

and indeed any such factor can be factored like this using the above trick:

$s^2-a=s^2-(\sqrt{a})^2=(s-\sqrt{a})(s+\sqrt{a})$ .

You asked about other ways.

2. The other way is via the Factor Theorem:

$k$ a root $\Leftrightarrow$ $(s-k)$ a factor,

therefore we look for the roots of $s^2-0.01$ i.e. we solve it equal to zero:

$s^2-0.01=0\Rightarrow s^2=0.01\Rightarrow s=\pm \sqrt{0.01}=\pm 0.1$ .

Therefore we have roots $+0.1$ and $-0.1$ and so factors $(s-0.1)$ and $(s+0.1)$ :

$\Rightarrow s^2-0.01=(s-0.1)(s+0.1)$ .

3. The same except that at $s^2-0.01$ we use the ‘-b’ formula with $a=1,\,b=0,\,c=-0.01$ :

$\displaystyle s=\frac{-0\pm\sqrt{0^2-4(1)(-0.01)}}{2}=\frac{\pm 0.2}{2}=\pm0.1$ ,

and the rest follows similarly.

4. Pragmatically, you have $s^2-0.01$ but the answer has $(s-0.1)(s+0.1)$ . Multiply this out:

$s^2+0.1s-0.1s-0.01=s^2-0.01$ ,

so they are equal.

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MATH7021: Spring 2018, Week 12

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MATH7021: Spring 2018, Week 12

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Week 12

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