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# Core Maths 4: Differential equations Watch

1. Hi there

Please could someone take a loko at the below and for question 4, tell me why I am getting a slightly different answer to whats in the book, and for question 5, explain how to do this as I have completely gone off the rails lol

I am really struggling with these types of questions, so if anyone can give me a little Differential equations for dummies guide or point me in the direction of one, I would be most grateful, cheers;

Question 4:

Question 5:

Many thanks
2. First one: How on Earth are you getting exponentials? Separate the variables and integrate!
3. first one:

integral 1/h^1/2 dh = integral k dt

where do you get exp from?
4. Second one is no better.

Factorise the denominator of the integrand and split it into partial fractions.
5. 1/(x(x-1) does integrate to ln(x(x-1)

you need to split it into partial fractions.
6. lol,there is an echo is this thread
7. And I can only echo too

OP you seem to have misunderstood this topic completely

You do not even begin correctly

8. (Original post by TenOfThem)
And I can only echo too

OP you seem to have misunderstood this topic completely

You do not even begin correctly

I ignored that as the negative sign can be incorporated within the constant of proportionality. Think the other problems are more glaring!
9. (Original post by Mr M)
Think the other problems are more glaring!
Agreed
10. ok thank you all for your replies.

I have managed to solve question 5, but am still having trouble with question 4.

With question 4, is this going in the right direction;

2√h = -kt + c

4h = (-kt)² + c

h = 1/4(-kt)² + c

???

Then does, c = 200?
11. (Original post by jackie11)
ok thank you all for your replies.

I have managed to solve question 5, but am still having trouble with question 4.

With question 4, is this going in the right direction;

2√h = -kt + c

4h = (-kt)² + c

h = 1/4(-kt)² + c

???

Then does, c = 200?
12. (Original post by jackie11)
ok thank you all for your replies.

I have managed to solve question 5, but am still having trouble with question 4.

With question 4, is this going in the right direction;

2√h = -kt + c

4h = (-kt)² + c

h = 1/4(-kt)² + c

???

Then does, c = 200?
Definitely going in the right direction, but I'm not convinced on your workings between lines 1 and 2 as you should square all of the right hand side as well. I know that a constant squared is just another constant, but you should also get -2ckt which you haven't got. Therefore, I would just substitute some values into the top line to find values for k and c. I think that works out fine.
13. (Original post by Quip)
Definitely going in the right direction, but I'm not convinced on your workings between lines 1 and 2 as you should square all of the right hand side as well. I know that a constant squared is just another constant, but you should also get -2ckt which you haven't got. Therefore, I would just substitute some values into the top line to find values for k and c. I think that works out fine.
ok how about this;

2√h = -kt + c

when h = 200, t = 0

2√200 = c

when h = 128, t = 5

2√128 = -5k + 2√200
2√128 - 2√200 = -5k
-5.656 = -5k
5.656 = 5k
1.13 = k
14. (Original post by jackie11)
ok how about this;

2√h = -kt + c

when h = 200, t = 0

2√200 = c

when h = 128, t = 5

2√128 = -5k + 2√200
2√128 - 2√200 = -5k
-5.656 = -5k
5.656 = 5k
1.13 = k
If this is still relevant, yes that's right (I think), but it may be more useful to keep k as a surd though.

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Updated: July 26, 2012
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