Hello i need help with this differential equations system!

x,u and l are functions of t... so u(t), x(t) and l(t)

tf is free.

Could you find the solutions?

boundary conditions

x(0)=1

x(tf)=0

l(tf)=2

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- May 16th 2010, 08:58 AMmagodiafanoDifferential first order equations
Hello i need help with this differential equations system!

x,u and l are functions of t... so u(t), x(t) and l(t)

tf is free.

Could you find the solutions?

boundary conditions

x(0)=1

x(tf)=0

l(tf)=2 - May 17th 2010, 05:12 AMchisigma
First step is to 'attack' ...

(1)

... that contains only the 'unknown function' l. Its solution is...

(2)

Now from the third equation we obtain...

(3)

... and if we insert (3) in the first equation we obtain...

(4)

... the solution of which is...

(5)

The values of and are obtained from the 'initial conditions'...

Kind regards

- May 17th 2010, 05:28 AMmagodiafano
uhm.. there is something that is not clear (sorry...):

1) why did you use

ibstead of ?? perhaps in order to simplify the calculations after?

2)

it's not clear the second and the third term of the result...

why is it not:

???

besides, since it is a definite integral between t0 and tf, what's the final result using this considerations? - May 17th 2010, 06:11 AMchisigma
I apologize for some mistakes caused by my hurry (Headbang) ...

The DE ...

(1)

... has solution...

(2)

... where...

(3)

Today my 'computation capability' is not excellent... sorry again (Worried) ...

Kind regards

- May 17th 2010, 07:20 AMmagodiafano
uhm i don't understand why do you do the multiplication and not the sum of the terms?

i think that result should be:

i think that i can consider each term per time and then do the sum of the results... is it right? - May 17th 2010, 10:47 AMmagodiafano
is there somebody could help me? :(

- May 17th 2010, 11:08 AMchisigma
- May 18th 2010, 09:33 AMmagodiafano
Now it's clear...

but i've another question for you!

I have three boundary conditions.. one about l(t), with whom i can easily find C1... and the other two about x(t)! the problem is if i use the boundary condition for t=0 (so x(0) ), i obtain that the condition on x(1) is not satisfied unless i fix tf (final istant)... is should be right?