The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Core Maths 4: Differential equations

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 :smile:
First one: How on Earth are you getting exponentials? Separate the variables and integrate!
(edited 11 years ago)
Reply 2
Avatar for mathz
mathz
8
first one:

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

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

∫1x−x2 dx≠−ln⁡∣x−x2∣+c\int \frac{1}{x-x^2} \, dx \neq - \ln |x-x^2| + c

Factorise the denominator of the integrand and split it into partial fractions.
Reply 4
Avatar for mathz
mathz
8
1/(x(x-1) does integrate to ln(x(x-1)

you need to split it into partial fractions.
Reply 5
Avatar for mathz
mathz
8
lol,there is an echo is this thread
Reply 6
Avatar for TenOfThem
TenOfThem
18
And I can only echo too

OP you seem to have misunderstood this topic completely

You do not even begin correctly

dhdt=−kh\frac{dh}{dt} = -k\sqrt{h}
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

dhdt=−kh\frac{dh}{dt} = -k\sqrt{h}


I ignored that as the negative sign can be incorporated within the constant of proportionality. Think the other problems are more glaring!
Reply 8
Avatar for TenOfThem
TenOfThem
18
Original post by Mr M
Think the other problems are more glaring!


Agreed
Reply 9
Avatar for jackie11
jackie11
OP
4
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?
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?


(a+b)2≠a2+b2(a + b)^2 \neq a^2 + b^2
Reply 11
Avatar for Quip
Quip
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.
Reply 12
Avatar for jackie11
jackie11
OP
4
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
Reply 13
Avatar for Quip
Quip
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.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you