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DGL in Mathematica

Posted 10 years ago

Dear Community,

i am new in Mathematica and want to solve an Equation like this: dx/dt+a*x=b

I know that there ist the function "Solve", but I tried this "Solve[dx/dt+a*x=b, x]" and nothing happens. Not even an error was shown. So i do not know how to describe the term dx/dt.

For more infos: x(t) is a time sequence with 150 datas. Like X=[3,4,0,2,3,1,....]. In Wolfram Alpha I could just solve this equation. The Result was: x(t) = b/a+c_1 e^(-a t).

Is it right that i "just" need to define dx/dt as x'? Even if I don't have a "normal" function which i can define?

Thank you all!

Regards, Julia

POSTED BY: Julia Telge
8 Replies
POSTED BY: Sean Clarke

That right hand side term $\sum_{i=2}^\infty b_i x_i[t]$ is a force term or perturbation term. Call it F[t] and the equation is

DSolve[{a x[t] + g x'[t] + x''[t] == F[t]}, x[t], t]

with solution (the case F = const was a test)

enter image description here

If you have initial conditions the solution might become simpler when the initial conditions are met.

Then you should clarify whether you consider a continuous $t \in \mathbb{R}$ or a discrete $t \in \mathbb{N}$.

POSTED BY: Udo Krause
Posted 10 years ago
POSTED BY: Julia Telge

Na, Udo hat das doch alles schön gelöst.

Perhaps everything gets clearer when you substitue the K[ i ] by other variables:

DSolve[a x[t] + g x'[t] + x''[t] == F[t], x[t], t] /.   K[1] -> \[Xi] /. K[2] -> \[Eta]

But then remains the question what do you mean by xi[ t ]? They seem to be discrete values, which are to be represented by a function F[ t ].

I understand it like this: to each t ( the specific t valid for xi ) you go to calculate

A[t] = Sum[ x[i, t], {i, 2, n}]

and you will get a set of A - values (for the given t's) representing the value of F for that t. And now you construct an Interpolation - Function (see Interpolation in the help-menu) which you could pass to the solution of your DGL.

But I am afraid the Integration in this case will only work numerically.

Regards Hans

POSTED BY: Hans Dolhaine
Posted 10 years ago

Hallo Hans,

danke für die schnelle Antwort.

Ich würde dort also die Summe erneut einsetzen und anstatt [t] dann [K[1]], oder? Meinst du das damit? Das heißt, um das explizit lösen zu können, muss ich dann ebenfalls das Integral berechnen, nicht wahr?

Puh, ich habe mir das irgendwie schöner vogestellt. :D

VG Julia

POSTED BY: Julia Telge

Die C[ i ] sind Integrationskonstanten, die K[ i ] neue Namen für die Variablen in der Integration.

POSTED BY: Hans Dolhaine
Posted 10 years ago

Thank you for your reply. In the second Output, i do not understand waht the K[1], K[2] stands for. Furthermore I do not know what F[K[1]] stands for. What should i interpret? I have F[t]=$\sum{i=2}^\infty bi x_i[t]$ . What is F[K[1]]?

I have a discrete t. The xi[t] are time series like x1=[3,6,8,9,...,n], x_2=[6,9,10,12,...,n] etc. for an example.

Regards Julia

POSTED BY: Julia Telge
Posted 10 years ago

Thank you for your answer. Sorry, that i reply that late.

Now, I was working and trying the new command DSolve and I think,now I understand it. So i was trying something new. I have to solve this monster: x1''[t]+gx1'[t]+a*x1=sum(i=2)^N bixi,x).

xi[t] are different time series. For example x1=[1,2,3,4,...,n], x2=[4,6,3,5,6,....,n] and so on. I have N time series and at the left side of the equation there is only time series x1. On the other side i want to have something like that: b2x2+b3x3+... and so on. So the equation have the character as described in the top.

I tried it in Mathematica: Input: DSolve[{a x[t] + g x'[t] + x''[t] == Sum_ (i = 0)^infinity bi*xi}, x[t], t] but the program replied with

"The function x appears with no arguments. Outpu: DSolve[{a x[t] + g Derivative[1][x][t] + (x^[Prime][Prime])[t] == 0^infinity (b : Blank[0]) Sum_ (x : Blank[0])}, x[t], t]"

So i think that something went wrong with the sum. How can I solve this problem?

Thanks at all!

Regards,Julia

POSTED BY: Julia Telge
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