Differential first order equations

Apr 2010
9
0
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?

\(\displaystyle

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

\(\displaystyle
\frac{dl}{dt}=l
\)

\(\displaystyle
l -2u-2=0


\)

boundary conditions
x(0)=1
x(tf)=0
l(tf)=2
 
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chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
First step is to 'attack' ...

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

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

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

Now from the third equation we obtain...

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

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

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

... the solution of which is...

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

The values of \(\displaystyle c_{1}\) and \(\displaystyle c_{2} \) are obtained from the 'initial conditions'...

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
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Apr 2010
9
0
uhm.. there is something that is not clear (sorry...):

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

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

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

???

besides, since it is a definite integral between t0 and tf, what's the final result using this considerations?
 

chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
I apologize for some mistakes caused by my hurry (Headbang) ...

The DE ...

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

... has solution...

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

... where...

\(\displaystyle a(t) = -1\)

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

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

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
Apr 2010
9
0
uhm i don't understand why do you do the multiplication and not the sum of the terms?

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

i think that result should be:

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




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

chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
uhm i don't understand why do you do the multiplication and not the sum of the terms?

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

i think that result should be:

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

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

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

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

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

In fact is...

\(\displaystyle \frac{dx}{dt}= - c_{2} e^{-t} + \frac{c_{1}}{2} e^{t}\)

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

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
Apr 2010
9
0
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?
 
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