**I am emailing a link of this to everyone on the class list every week. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jpmccarthymaths@gmail.com and I will add you to the mailing list.**

## Maple Test

I have given ye a sample Maple Testt and ye will have your Maple Test in Week 12.

## Study

Please feel free to ask me questions via email or even better on this webpage — especially those of us who struggled in the test.

Please find a reference for some of the prerequisite material here.

## Week 10

We continued our study of Laplace Methods with a review of partial fractions. We also spoke about the inverse of the Laplace transform. I also gave out solutions of the Winter 2012 Laplace questions so that ye can see how I mark these questions.

## Weeks 11 & 12

We continue our work on Laplace Methods and see how they can solve differential equations.

## Week 13

We will do the exam paper from Summer 2012.

## Math.Stack Exchange

If you find yourself stuck and for some reason feel unable to ask me the question you could do worse than go to the excellent site math.stackexchange.com. If you are nice and polite, and show due deference to these principles you will find that your questions are answered promptly. For example this rather technical question about the inverse Laplace transform.

## 2 comments

Comments feed for this article

April 25, 2013 at 9:06 am

Student 35JP,

Using all the steps as per some of your examples I worked out Question 2 (iv). The answer I got was:

from

Can you let me know if this is close?

April 25, 2013 at 9:18 am

J.P. McCarthyYou factorised the bottom incorrectly which led to a simplified solution. If the bottom was then your method would have worked. We give an answer with on top and then one with .

1.

Starting with we note that this isn’t in the tables so perhaps we need to write it as a sum of things that are in the table and this means do a partial fraction expansion.

The first step here is to factorise the denominator:

It appears that we can’t factorise so we look to see does it have any roots using the formula: no real roots means no real factors. We examine

hence we have complex roots and hence no (real) factors.

This is not ideal but we can rescue the situation if we note that there is a similarity

.

In particular if we could complete the square

,

then we might be able to write our transform as the sum of a shifted sine and a shifted cosine.

So we do what we want to do

.

Hence we want and . Hence we have

.

Now this is a nice situation because this is exactly a shifted

,

so comes back with a shift of :

.

2.

Everything is the same down until we get

.

Now at this point we should say that perhaps we can write this as a sum of shifted cosines and sines:

.

First we make the shifted sine happen by getting an in the numerator:

.

Now we can split this

.

The first term is just the shifted cosine that we saw above while the second can be made into a shifted sine by taking out the four:

.

Regards,

J.P.