Problems with this integral

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  • Thread starter MikeSv
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  • #1
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Hello everyone.

Iam trying to get my head around a solution for an integral but I can't figure out how its done.

I have given the following :

x1'(t) = 0
x2'(t) =tx1(t)

Where " ' " indicates the derivative.

Talking the time integral the result is given by:

x1(t) = x1(t0)
x2(t) = x2(t0) + 1/2(t^2-t0^2)x1(t0)

It would be great if anyone could help me out or give me a hint.

Cheers,

Mike
 

Answers and Replies

  • #2
35,139
6,892
Hello everyone.

Iam trying to get my head around a solution for an integral but I can't figure out how its done.

I have given the following :

x1'(t) = 0
x2'(t) =tx1(t)

Where " ' " indicates the derivative.

Talking the time integral the result is given by:

x1(t) = x1(t0)
x2(t) = x2(t0) + 1/2(t^2-t0^2)x1(t0)

It would be great if anyone could help me out or give me a hint.

Cheers,

Mike
Your solution looks fine to me, although it's more complicated than it needs to be with all the subscripts.
For simplicity in writing, I'm going to rephrase your problem:

x' = 0
y' = tx

Here, both x and y are functions of t.

Since x' = 0, then ##x = k_1##, for some constant ##k_1##. After substitution into the second equation, you get
##y = \frac 1 2 k_1t^2 + k_2##
It's always a good idea to verify that your solution actually works, by substituting back into the original system of equations.

Since there are no initial conditions given (or at least shown here), we're done.

Note that this is a very simple system of differential equations, one that can be "uncoupled" by substitution. More complicated systems, in which each derivative is in terms of the other function, require much more complicated techniques.
 
  • #3
35
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Hi and thanks for the reply


I guess I just got confused by the subsribts.
And if I see correctly the integral is evaluated from t0 to t.

Cheers,

Mike
 
  • #4
35,139
6,892
I guess I just got confused by the subsribts.
And if I see correctly the integral is evaluated from t0 to t.
No need to use definite integrals.
##x'(t) = 0 \Rightarrow x(t) = \int 0 dt = k_1##
##y'(t) = t k_1 \Rightarrow y(t) = k_1 \int t~ dt = \frac 1 2 k_1 t^2 + k_2##

Of course, since the problem is in terms of x1 and x2, the solutions should be as well.
 
  • #5
35
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Thank you so much for clarifying!
This helped a lot,!

Cheers,

Mike
 

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