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Degree: Mathematics (Bachelors) - Birmingham University
Hi! My name is Ed Wong and I'm a third year Mathematics student at the University of Birmingham. I have a great passion for maths, and would love to tutor anyone who is finding it difficult. I would say I am great at explaining my solutions to anyone- no matter the mathematical ability!
I achieved A* in GCSE maths, and A* in both Mathematics and Further Mathematics at A level.
Whether if you want to just pass your A-Levels, or even go on to further study, I will ensure that you have an enjoyable time while you learn with me.
I look forward to meeting you!
|Maths||A Level||£20 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
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Lisa (Parent) July 21 2016
Bryce (Student) July 19 2016
Lisa (Parent) July 18 2016
Bryce (Student) July 21 2016
When solving any differential equation, the first method to consider is the seperation of variables. This is the simplest method and, conveniently, it works in this case. To seperate variables:
1. Put all of the same type of variables on their own side. In this case, our two variables are t and θ. So in this case, divide everything by (120-θ) and multiply everything by dt. This leaves us with dθ/(120-θ)=λdt.
2. The next step is to integrate both sides.
On the left side, we have a variable ^-1 so our answer must have a natural log of the denominator. So lets work backwards, let's assume our answer is ln(120-θ). If we differentiate this, this gives us -1/(120-θ), which is what we started with, but multiplied by -1. Therefore, we must have -ln(120-θ).
On the right side, this is simply integrating a constant (as λ is a constant), so we have λt +c. REMEMBER TO ADD OUR CONSTANT OF INTEGRATION!
This leaves us with: -ln(120-θ)=λt +c.
3. Next we find what c is. This is found by applying the initial conditions given in the question, i.e. θ=20, t=0. Plugging this in and rearranging for c, we have c=-ln(100), leaving us with: -ln(120-θ)=λt-ln(100).
4. We now have our final expression, but it isn't in the correct form. Therefore, we must use the rules of exponentials to manipulate it. This will leave us with the correct form of: