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Degree: MSci Theoretical Physics (Masters) - Birmingham University
Hi, I'm Dom, a third year Theoretical Physics student at the University of Birmingham. I have always been fascinated by the rich subjects that are maths and physics and look to convey that enthusiasm in my teaching.
I have previous experience as a tutor and employ a friendly and patient approach, where I am willing to explain concepts in many different ways to ensure you improve your understanding. I appreciate that everyone learns differently and will tailor my teaching style to you, so that you get the most out of the tutorials.
What the sessions consist of is entirely up to you and depends on what you would find most useful. Let me know what exam board you are on and the general area you wish to cover in the tutorial and I will ensure that I am completely up to date with your specific course. Then, as well as improving your fundamental understanding of the topic, I can bring some example questions for us to go through.
It is well known that in modern day examinations, exam technique is a considerable factor in achieving excellent marks. As such I am more than happy for sessions to be focused on this and we can go through past papers to ensure that you are well prepared for whatever the examiners may throw at you!
For more information about the sessions or anything else, please don’t hesitate to send me a message or book a free ‘Meet the Tutor’ session.
|Further Mathematics||A Level||£20 /hr|
|Maths||A Level||£20 /hr|
|Physics||A Level||£20 /hr|
|Further Mathematics||GCSE||£18 /hr|
|Human Biology (AS)||A-Level||A|
|Before 12pm||12pm - 5pm||After 5pm|
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This is a question taken from a core 4 paper and is a typical example of a differential equation question.
The first thing to notice about this equation is that it is "separable", meaning we can rearrange it to get
e^(2y) dy = - tan(x) dx
Now we can solve this by integrating both sides. We know how to integrate the left hand side, and we get (1/2)e^(2y), but how can we integrate -tan(x)?
To see how we can do this, we write
-tan(x) = -sin(x) / cos(x)
Then, we realise that the numerator is the derivative of the denominator, and so integrating -tan(x) gives ln(|cos(x)|) + C, where C is the constant of integration.
So, we now have that
(1/2)e^(2y) = ln(|cos(x)|) + C
Now we apply the condition that y(x=0) = 0, giving
1/2 = C
Subbing this in, we have
(1/2)e^(2y) = ln(|cos(x)|) + 1/2
e^(2y) = 2ln(|cos(x)|) + 1
The question asked for the answer to be written in the form y = f(x), and so we need to get the y out of the exponent, which we can do by taking ln of both sides to give
2y = ln( 1 + 2ln(|cos(x)|) )
And so the final answer is
y = (1/2) ln( 1 + 2ln(|cos(x)|) )see more