# Differential first order equations

• May 16th 2010, 08:58 AM
magodiafano
Differential 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?

$

\frac{dx}{dt}=-x+u
$

$
\frac{dl}{dt}=l
$

$
l -2u-2=0

$

boundary conditions
x(0)=1
x(tf)=0
l(tf)=2
• May 17th 2010, 05:12 AM
chisigma
First step is to 'attack' ...

$\frac{dl}{dt} = l$ (1)

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

$l(t)=2 c_{1} e^{t}$ (2)

Now from the third equation we obtain...

$u(t) = c_{1} e^{t} - 1$ (3)

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

$\frac{dx}{dt} = -x + c_{1} e^{t} - 1$ (4)

... the solution of which is...

$x(t) = c_{2} e^{-t} + \frac{c_{1}}{2} e^{t} - 1$ (5)

The values of $c_{1}$ and $c_{2}$ are obtained from the 'initial conditions'...

Kind regards

$\chi$ $\sigma$
• May 17th 2010, 05:28 AM
magodiafano
uhm.. there is something that is not clear (sorry...):

1) why did you use
$2 c_{1}$
ibstead of $c_{1}$ ?? perhaps in order to simplify the calculations after?

2)
$x(t) = c_{2} e^{-t} + c_{1} e^{2 t} - e^{t}$

it's not clear the second and the third term of the result...
why is it not:
$x(t) = c_{2} e^{-t} + c_{1} e^{t} - t$

???

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

The DE ...

$\frac{dx}{dt} = - x + c_{1} e^{t} - 1$ (1)

... has solution...

$x (t) = e^{\int a(t) dt} \cdot \{b(t)\cdot e^{-\int a(t) dt} dt + c_{2}\}$ (2)

... where...

$a(t) = -1$

$b(t) = c_{1} e^{t} - 1$ (3)

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

Kind regards

$\chi$ $\sigma$
• May 17th 2010, 07:20 AM
magodiafano
uhm i don't understand why do you do the multiplication and not the sum of the terms?

$
\frac{dx}{dt} = - x + c_{1} e^{t} - 1
$

i think that result should be:

$
x (t) = e^{-t} + c_{1} e^{-t} - t + c2
$

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 AM
magodiafano
is there somebody could help me? :(
• May 17th 2010, 11:08 AM
chisigma
Quote:

Originally Posted by magodiafano
uhm i don't understand why do you do the multiplication and not the sum of the terms?

$
\frac{dx}{dt} = - x + c_{1} e^{t} - 1
$

i think that result should be:

$
x (t) = e^{-t} + c_{1} e^{-t} - t + c2
$

I think that i can consider each term per time and then do the sum of the results... is it right?

For the DE...

$\frac{dx}{dt} = - x + c_{1} e^{t} -1$ (1)

... it is easy to verify that the solution is...

$x(t) = c_{2} e^{-t} + \frac{c_{1}}{2} e ^{t} -1$ (2)

In fact is...

$\frac{dx}{dt}= - c_{2} e^{-t} + \frac{c_{1}}{2} e^{t}$

$- x + c_{1} e^{t} -1 = -c_{2} e^{-t} + \frac{c_{1}}{2} e^{t}$ (3)

Kind regards

$\chi$ $\sigma$
• May 18th 2010, 09:33 AM
magodiafano
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?