• We are pleased to announce that the winner of our Feedback Prize Draw for the Winter 2024-25 session and winning £150 of gift vouchers is Zhao Liang Tay. Congratulations to Zhao Liang. If you fancy winning £150 worth of gift vouchers (from a major UK store) for the Summer 2025 exam sitting for just a few minutes of your time throughout the session, please see our website at https://www.acted.co.uk/further-info.html?pat=feedback#feedback-prize for more information on how you can make sure your name is included in the draw at the end of the session.
  • Please be advised that the SP1, SP5 and SP7 X1 deadline is the 14th July and not the 17th June as first stated. Please accept out apologies for any confusion caused.

Solving differential equations

O

Oldy-but-Goody

Member
Probably a silly question this, but my maths isn't what it once was and I'm currently figuring out differential equations. The thing that stumping me a bit is introducing a constant of integration.......I'm not sure which side of the equation it needs to be added to! I'm sort of thinking it can be added to either side but I'm not confident enough in my reasoning to be sure.......can anybody possibly help me with this?

Cheers.
 
Oldy-but-Goody said:
Probably a silly question this, but my maths isn't what it once was and I'm currently figuring out differential equations. The thing that stumping me a bit is introducing a constant of integration.......I'm not sure which side of the equation it needs to be added to! I'm sort of thinking it can be added to either side but I'm not confident enough in my reasoning to be sure.......can anybody possibly help me with this?

Cheers.
Click to expand...

It can be added to either side. Adding it to one side is equivalent to adding the negative of it to the other.

e.g.

f'(x)=x^2

then f(x)+c=(x^3)/3

or f(x)=(x^3)/3+d

Where d and c are constants with d=-c.
 
Oldy-but-Goody said:
Probably a silly question this, but my maths isn't what it once was and I'm currently figuring out differential equations. The thing that stumping me a bit is introducing a constant of integration.......I'm not sure which side of the equation it needs to be added to! I'm sort of thinking it can be added to either side but I'm not confident enough in my reasoning to be sure.......can anybody possibly help me with this?

Cheers.
Click to expand...

Thanks!
 
OK, possibly another dumb question. My first question was prompted whilst looking at the separation method, I understand that now and seem to be getting derivations correct. However, I've started on the integrating factor method and am stumped again! I'm working through examples in both a 10 year old copy of FAC & the current CT4 notes, in CT4 it says:

'NB we don't have to bother about the constant of integration here. The constants will cancel out when we multiply every term by the integrating factor'

However, in the example in FAC the solution does include c and it says:

'c is an arbitrary constant'

I'm a bit confused, the first example had me thinking that the integrating factor method doesn't require the constant when integrating, but then the second seems to contradict this? Not sure what I'm missing and if there is something special about the second example which specifically requires a c term?
 
Hi Oddy-but-Goody,

Separation of variables is in fact a simplified version of the integrating factor technique so you've already been applying it successfully.

The constant in the integrating factor bit can safely be ignored because you're only going to apply another constant later and use the boundary conditions of the problem you're looking at.

To see this, why not try it both ways?

1. Have a constant C floating about, to be joined later by another constant K, say. You will need to set C + K to be something so that the boundary condition is satisfied. In CT4 this is usually either that the probability of going somewhere in no time at all is 0 or that the probability of staying where you are if you've no time to go anywhere is 1.

2. Don't worry about the C but later there will be a constant K, set so that the boundary condition is satisfied.

It really doesn't matter whether you set C + K to be something or set K to be something,

Good luck!
John
 
John Potter said:
Hi Oddy-but-Goody,

Separation of variables is in fact a simplified version of the integrating factor technique so you've already been applying it successfully.

The constant in the integrating factor bit can safely be ignored because you're only going to apply another constant later and use the boundary conditions of the problem you're looking at.

To see this, why not try it both ways?

1. Have a constant C floating about, to be joined later by another constant K, say. You will need to set C + K to be something so that the boundary condition is satisfied. In CT4 this is usually either that the probability of going somewhere in no time at all is 0 or that the probability of staying where you are if you've no time to go anywhere is 1.

2. Don't worry about the C but later there will be a constant K, set so that the boundary condition is satisfied.

It really doesn't matter whether you set C + K to be something or set K to be something,

Good luck!
John
Click to expand...

Thanks......it all makes sense now!
 
You must log in or register to reply here.
Back
Top