You are Here: Home >< Maths

# Differential equations questions? Watch

1. Hi, I've been working on some C4 differential equation questions and I have a few questions. I have attached a scan of my workings and a copy of the questions.

For Q12, I am able to solve most of the question, but I'm unsure how you would "show that this model satisfies the differential equation dh/dt = -k(2h-h^2)^1/2. Do you just insert h=0 and dh/dt=0?

For Q14, I'm not sure how to answer b), surely if you only use N=100 and t=0, then you would get a straight line graph? Also, for part c, my answer is slightly different to what they have. Mine is 1/500ln(n/[4(500-n)]) = 1/5000t whereas in the textbook it says 1/500ln(4n/500-n)=1/5000t. I think their answer is different as they placed the constant on the left rather than on the right (like I did), so they got a different value for c. But as I have a different value of c, it's completely wrong?

For Q15, for b) I'm not sure where I substitute -1 into. And I don't really understand what c/d are saying.

2. (Original post by Bazinga?)
.
For q12, I've arrived at an inelegant solution. First consider differentiate and then manipulate the original equation to attain the given result.
3. (Original post by joostan)
For q12, I've arrived at an inelegant solution. First consider differentiate and then manipulate the original equation to attain the given result.
What do you mean by an inelegant solution?

So when you integrate h=1-sin(kt) then you get dh/dt = -cos(kt). This is probably wrong, but do you equate the dh/dt's? So you get -cos(kt) = -h^1/2, so cos(kt) = h^1/2. Then can you substitute h=(1-1/2t)^2? However if I do this, I lose the h's and everything is in terms of t?
4. (Original post by Bazinga?)
What do you mean by an inelegant solution?

So when you integrate h=1-sin(kt) then you get dh/dt = -cos(kt). This is probably wrong, but do you equate the dh/dt's? So you get -cos(kt) = -h^1/2, so cos(kt) = h^1/2. Then can you substitute h=(1-1/2t)^2? However if I do this, I lose the h's and everything is in terms of t?
Inelegant as in there's probably an easier way.
Spoiler:
Show
Chain rule.

When I said original equation I meant rewrite
Spoiler:
Show
Consider how sin and cos are related.
5. (Original post by Bazinga?)
Hi, I've been working on some C4 differential equation questions and I have a few questions. I have attached a scan of my workings and a copy of the questions.

For Q12, I am able to solve most of the question, but I'm unsure how you would "show that this model satisfies the differential equation dh/dt = -k(2h-h^2)^1/2. Do you just insert h=0 and dh/dt=0?
You need to insert h=1-sinkt in both side and to show the equation
For Q14, I'm not sure how to answer b), surely if you only use N=100 and t=0, then you would get a straight line graph? Also, for part c, my answer is slightly different to what they have. Mine is 1/500ln(n/[4(500-n)]) = 1/5000t whereas in the textbook it says 1/500ln(4n/500-n)=1/5000t. I think their answer is different as they placed the constant on the left rather than on the right (like I did), so they got a different value for c. But as I have a different value of c, it's completely wrong?
For a) sub N=100 in the RHS you will get 8 for dN/dt as rate
for b) Yes rhis the gradient line of the curve for N=constat
fe. dN/dt=8 (for N=100) -> N=8t+c -< this c is 100 for t=0 => N=8t+100
for other N=const you will get other lines. These lines cover the curve
for c)
1/(N(500-N))=A/N+B/(500-N)->(B-A)=0 and 500A=1 ->A=B and A=1/500
So 1/500 (1/N+1/(500-N))dN=1/5000 dt
from this you will get ln(N*(500-N)=1/10*t+C
For Q15, for b) I'm not sure where I substitute -1 into. And I don't really understand what c/d are saying.
you need to substitute y=1/4(x+1)^2 into dy/dx=Vy and to show the equation
y>=0 because of Vy -> x>-1 after differentiation
6. (Original post by joostan)
Inelegant as in there's probably an easier way.
Spoiler:
Show
Chain rule.

When I said original equation I meant rewrite
Spoiler:
Show
Consider how sin and cos are related.
Ah yes, I thought I misinterpreted what you originally said. Thanks again for your help!

(Original post by ztibor)
You need to insert h=1-sinkt in both side and to show the equation

For a) sub N=100 in the RHS you will get 8 for dN/dt as rate
for b) Yes rhis the gradient line of the curve for N=constat
fe. dN/dt=8 (for N=100) -> N=8t+c -< this c is 100 for t=0 => N=8t+100
for other N=const you will get other lines. These lines cover the curve
for c)
1/(N(500-N))=A/N+B/(500-N)->(B-A)=0 and 500A=1 ->A=B and A=1/500
So 1/500 (1/N+1/(500-N))dN=1/5000 dt
from this you will get ln(N*(500-N)=1/10*t+C
This is what I got, and you work out c to be 1/500ln0.25. However this is not true (well it's wrong according to the answers).
you need to substitute y=1/4(x+1)^2 into dy/dx=Vy and to show the equation
y>=0 because of Vy -> x>-1 after differentiation
I've substituted y=1/4(x+1)^2 into the dy/dx=y^1/2, but I don't really follow what I should do next.
7. (Original post by Bazinga?)
Ah yes, I thought I misinterpreted what you originally said. Thanks again for your help!

This is what I got, and you work out c to be 1/500ln0.25. However this is not true (well it's wrong according to the answers).
I wrote

I've substituted y=1/4(x+1)^2 into the dy/dx=y^1/2, but I don't really follow what I should do next.[/QUOTE]

is nonnegative by definition
and for real set
8. (Original post by ztibor)
I wrote

I've substituted y=1/4(x+1)^2 into the dy/dx=y^1/2, but I don't really follow what I should do next.

is nonnegative by definition
and for real set[/QUOTE]

For 14, I understand what you've done, but why can't I do what I did? I don't see where i hav gone wrong. And with what you're saying 1/500lnN(500-N) = 1/10t + c, when t=0 and N=100, then ln100(400) = c, so c = ln40000?

I understand 15 now, thanks
9. (Original post by Bazinga?)

is nonnegative by definition
and for real set
For 14, I understand what you've done, but why can't I do what I did? I don't see where i hav gone wrong. And with what you're saying 1/500lnN(500-N) = 1/10t + c, when t=0 and N=100, then ln100(400) = c, so c = ln40000?

I understand 15 now, thanks [/QUOTE]

I made a mistake so you are right
SO

so
c=1/4 or C=ln(1/4)
10. (Original post by ztibor)
For 14, I understand what you've done, but why can't I do what I did? I don't see where i hav gone wrong. And with what you're saying 1/500lnN(500-N) = 1/10t + c, when t=0 and N=100, then ln100(400) = c, so c = ln40000?

I understand 15 now, thanks
I made a mistake so you are right
SO

so
c=1/4 or C=ln(1/4)[/QUOTE]

Surely you'd sub in t=0 and N=100 to find c?

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: May 15, 2013
Today on TSR

### Does race play a role in Oxbridge admissions?

Or does it play no part?

### The worst opinion about food you'll ever hear

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.