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# Differential Eqns watch

1. I get the first equation down to:

How do I get rid of the mod signs (something to do with plus/minus?) and apply the initial conditions to find A?

Then how would I go about solving equation 2?
2. Can P be negative?
3. (Original post by JoMo1)
Can P be negative?
I suppose not from the context as it is an 'element' - a bit vague though.

So that would give

which means

How would I solve

?
4. (Original post by TheEd)
I suppose not from the context as it is an 'element' - a bit vague though.

So that would give

which means

How would I solve

?
use integrating factors
use integrating factors
I haven't learnt that yet so I can't do it but just in general, how would I get rid of a mod sign?

Would it have become:

?
6. (Original post by TheEd)
I haven't learnt that yet so I can't do it but just in general, how would I get rid of a mod sign?

Would it have become:

?
just absorb the signs into the constant A then applying boundary intial conidtions will determine it.
7. (Original post by TheEd)
I haven't learnt that yet so I can't do it but just in general, how would I get rid of a mod sign?

Would it have become:

?
take bQ to LHS

to solve the equation multiply it through by u(x) so that

LHS=w dQ/dt + bwQ

then look for w such that LHS=d(wQ)/dt

so use product rule and compare with wdQ/dt + bwQ to get a DE for w and solve that.
take bQ to LHS

to solve the equation multiply it through by u(x) so that

LHS=w dQ/dt + bwQ

then look for w such that LHS=d(wQ)/dt

so use product rule and compare with wdQ/dt + bwQ to get a DE for w and solve that.
So

where to now?
9. (Original post by TheEd)
So

where to now?
You can take A = 1 so w=exp(bt)
you don't need to bother with modulus signs for solns of DEs.

so LHS = d(wQ)/dt
RHS = wP

so d(wQ)/dt = wP so integrate both sides wrt t.
You can take A = 1 so w=exp(bt)
you don't need to bother with modulus signs for solns of DEs.

so LHS = d(wQ)/dt
RHS = wP

so d(wQ)/dt = wP so integrate both sides wrt t.
Where's RHS = wP come from?
11. (Original post by TheEd)
Where's RHS = wP come from?
dQ/dt = -bQ - dP/dt

but dP/dt = -aP

=> dQ/dt + bQ = aP

and then you multiplied the equation by w. Sorry I missed out the a in my last post.
dQ/dt = -bQ - dP/dt

but dP/dt = -aP

=> dQ/dt + bQ = aP

and then you multiplied the equation by w. Sorry I missed out the a in my last post.
So

EDIT: This is wrong as w is a function not a constant isn't it?
13. (Original post by TheEd)
So

EDIT: This is wrong as w is a function not a constant isn't it?
When you integrate the RHS, w is also a function of t/
When you integrate the RHS, w is also a function of t/
so does it have to be done by parts?

and assume .

Then .

For we then have .

Now let . Then the RHS converges to 0, implying that f is differentiable at 0 and .

--------

is differentiable on if the following limit exists for all fixed on the interval

With for :

15. (Original post by TheEd)
so does it have to be done by parts?
combine the powers of e
combine the powers of e
So

17. (Original post by TheEd)
So

That looks right to me.
That looks right to me.
But applying the initial condition that would imply

Wouldn't this imply that which can't be right as P has to initially be positive as it is "decaying" otherwise the system won't work?
19. (Original post by TheEd)
But applying the initial condition that would imply

Wouldn't this imply that which can't be right as P has to initially be positive as it is "decaying" otherwise the system won't work?
Just gone through it all myself and realised you've missed a constant of integration when you integrated awP so you should have an extra Cexp(-Bt) on the RHS of your soln
Just gone through it all myself and realised you've missed a constant of integration when you integrated awP so you should have an extra Cexp(-Bt) on the RHS of your soln
Ah yes it's because I didn't put in explicit integral signs so I forgot!

So it should be:

provided

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